Calculus 2 Final Review || Techniques of Integration, Sequences & Series, Parametric, Polar & More!

alright so in this video we are going to be doing some final review for calculus 2 okay so a couple things before we start here this review video is going to

consist of 30 questions just drawing from all topics for each problem definitely try it yourself first okay you can get a feel of what part of the problem you're struggling on in

particular okay and I'm gonna go through it right after you try it so you know you'll you can see if you happen to get it wrong which hopefully you don't but if you do you'll see where you got it

wrong and you'll see how to correct that there's also going to be a 100 question video linked in the description okay so if you want some more review after these

30 questions this 100 question video is an extended version of this talk to you final review okay I'm able to draw more questions per topic so if that's something that interests you if you

really want a ton of review I really encourage you to check that out lastly if you know if you've been watching my calc you videos if they've really been helping you I would really

appreciate it if you could consider supporting me on patreon it would definitely really help out and yeah I would definitely appreciate it so onto the first question

so for our first question we want to find the area bounded by the curves y equals 1 over X y equals x 1 equals 1/4 X and that's all for X greater than zero

okay so the first thing we need to do is obviously get an idea of what we're looking at here we need to graph this

okay so what let's graph our y equals 1 over X first that will just look like that right ok and make sure you label it so you're not integrating the wrong

thing when we actually have to go integrate later next we have y equals x we know what that looks like alright and I'm only drawing this for X is greater than 0 honestly you could

just kind of forget about this whole part right here so we have y equals x and then y equals 1/4 X is just going to

be y equals x but not as steep all right so that's y equals 1/4 X ok so now what area are we talking about here what's

that region bounded by the curves well that's right here okay that's this region right in here that's being bounded by all the curves okay it's

important that you don't that you identified the right region okay because if you don't if you're trying to find something like this right here that you know that's gonna throw you off completely so now we need to figure out

how we're going to actually find the area okay and of course we're gonna have to integrate now what you'll notice is that we have two separate top functions ok when we integrate between two curves

so we have a top function and a bottom function okay but here we have two separate top functions we have X at first and then we have 1 over X this that's gonna happen at that intersection

point right there okay that's where switches over so what we're gonna have to do is we can set up two separate integrals and the first integral will be from we know it's going to start at 0

okay goes three to zero we don't know whether this intersection point is so we got to figure that out but what is going to be our integrand well that's just going to be the top function minus the

bottom function okay function being X and the bottom function being 1/4 X so this will be X minus 1/4

X DX okay now what is going to be that an intersection point well what we can

do is set we can set X equal to 1 over X okay why are we doing that well we want to figure out where X crosses 1 over X

okay that's the idea so obviously here x squared equals 1 and we can take the square root of both sides X is going to equal plus or minus 1 okay and we're

dealing with positive X's here so X is going to equal 1 so that is our top bound hey that's your first that's your

first integral and the next is going to be from well you're going to start off at that one and you need to find this intersection for a year with the x-value

is there and that's going to give you your top bound but what is your integrand okay well your top function now is 1 over X so you're going to start off with 1 over X and then you can

subtract your bottom function which in this case is still 1/4 X okay now we can

find that top bound by setting 1/4 X equal to 1 over X so 1 over 4 X you can write that as x over 4 is equal to 1

over X now we can move things diagonally when it's two fractions equal to each other okay so this X can come up here this 4 can go up here hey and you can just multiply on both sides if you want

to see that okay basically across multiplication we get that x squared equals 4 okay take the square root of both sides x equals plus or minus 2 but

we don't care about the negative we care about the positive so x equals 2 that is our top bound now before we integrate here what we can

do is simplify this down a little bit so we have the integral from 0 to 1 of this is just well this is a

X minus a 1/4 X which is just a 3 over 4 X okay multiplied by 4 over 4 here and you'll see what I'm talking about if

that didn't make perfect sense now our next week we have this integral from 1 to 2 we don't need to simplify this at all ok this will be easier to

integrate by itself now when you integrate here you're going to get well this power will get raised to a 2 and then we'll divide by 2 so

we'll be getting a 3 over 8 x squared and that'll be evaluated from 1 to us from 0 to 1 and then we have plus we'll

do this in parentheses here this will be a will be under 'grill of 1 over X is a natural log absolute value of x we're only dealing with positive X's and our bounds so we don't really need to worry

about that absolute value so we can just write natural log of x next we'll have a minus well the integral of 1/4 X is

going to be x squared over 8 or 1 over 8 times x squared if that sounds better to you then this will be evaluated from 1

to 2 okay so now just doing our evaluation step here where our zero is gonna kill this off okay so we don't need to worry about the zero at all

we're just gonna have a 3/8 times 1 squared which is just going to be one full read anyway and then we'll have

over here a natural log up to minus 2 squared which is going to be 4 4 over 8 is 1/2 okay and that will be a minus a

natural log of 1 which is 0 okay that'll be minus a 1 squared which is 1 over 8

okay so simplifying this a little bit you get a 3/8 plus a natural log of Q minus a 1/2 you can see that we have

over 8 here ok so we're just gonna make this 1/2 of 4 over 8 so we'll have minus 4 over 8 and this will be calm plus plus and you

have plus warning okay simplifying this down a little bit more just combining your like terms here you have 3/8 plus 1/8 which is going to be

four eighths and then you have minus four east so that's going to go away and you're just left with natural log of 2 all right so one problem down twenty nine to go

are you so question number two we want to find the volume of the salad obtained by revolving the region bounded by y

equals 1 over X y equals 0 x equals 1 and x equals 4 about the x-axis okay so let's get an idea of what we're looking at here we have this region bounded by

these four curves here okay so we can graph that ok when you have 1 over X we

have y equals 0 which is going to be this line right here we have x equals 1

and we have x equals 4 so that's going to be we're talking about this region right here in blue well should color that in blue okay

and we're evolving that region about the x axis okay now the x axis of course that's right here so when we rotate it ok we get

so you can see here if you were to slice this solid this way your cross-sections would be circles and we know that the

area of a circle is PI R squared okay now if we want to find the volume of this solid what we're gonna do is add up

all of these cross-sectional areas okay and how do we do that well we take an

integral okay we integrate PI R squared okay and well how do we I mean how do we go about doing that well our bounds are

going to be from 1 to 4 okay and your radius well that's the distance from me get this blue out of the way and write

your radius the radius is from your curve to the axis of rotation okay so that is R now we know that that the

radius is 1 over X okay the distance from this curve 1 over X to the x axis is 1 over X okay so what we're going to

now have here is we're gonna replace this R with 1 over X and that's going to be getting squared okay so we can pull this pi out and of course we're gonna

have a DX on the end here we're gonna pull this pi out and then we view the integral from 1 to 4 of this will now be

a 1 over x squared DX okay now when you integrate 1 over x squared what you get

is negative 1 over X and that will be evaluated from 1 to 4 okay so now doing

that you get that volume is equal to pi times this will be a negative 1/4 okay

and that will be minus a negative 1 over 1 okay so it's basically negative 1/4

plus 1 which will be 3/4 so your answer is going to be that your equals 3 PI over 4 so just to recap here we started off by graphical that they

gave us and rotating about the x axis okay now we know that if we cut this solid this way our area is our cross

sections are going to be circles and the area of that is PI R squared okay so if we want to find the volume we add up all of those cross sectional areas okay and

we do that by integrating so we integrated we just I mean we integrated we evaluated and we found our answer

okay so as simple as that so question number three says we need to use the shell method to find the volume of the solid obtained by rotating the

region bounded by y equals x squared and y equals 4x minus x squared about the line x equals four okay so first you need to recall what the shell method

actually says okay the the formula for the shell method says that your volume

okay your volume on the wrong side here your volume is going to be equal to the integral from A to B of your

circumference okay so be your circumference times your height times

your width okay and we're getting these because remember by shell method we're talking about cylindrical shells okay so we're basically finding the volume of

a cylinder here that's kind of where this whole concept comes from okay so this is the formula for the shell method now let's see what our region is let's rotate it about the line x equals four

and I'm just gonna put a better picture of what we're doing here so first we have the graph y equals x squared okay

now this is something that we know what that looks like okay y equals x squared now next we have y equals 4x minus x squared okay now if

you don't know what that graph looks like cuz I don't know what that graph looks like off the top of my head you can just find the zeros okay you there's a negative x squared here so you know it's gonna be an upside-down parabola

but just find the zeros by setting y equal to zero you know just like in algebra and figuring out way of what X

values will make this Y zero okay so let's just do that here take out next we have 4 minus X left so you know that the zeros are going to be x equals 0 and x

equals 4 okay so those are our zeroes stay right here and right here okay this graph will

look like this and we're rotating about the line x equals four okay so what we're gonna get here is just this line

coming down and that's what we're rotating over all right now remember it's bounded by those two curves okay that's what a region is so that region

that we're talking about here is right here okay color and blue for a second okay we're rotating that around this line okay so now we're gonna have to set

up our integral and the first thing are we gonna be integrating with respect to X or with respect to Y well remember with cylindrical shells we are going to

be cutting parallel to the axis of rotation okay but with disk method with washer method its perpendicular so here alright our cuts are gonna look like

this okay and that's obviously integrating with respect to X think of Riemann sums okay so now let's set this

up we have the volume equaling the integral from what to what we know it starts at zero but where does our region end it ends at the intersection of x

squared and 4x minus x squared so what we need to do is set these equal to each

other okay so we can say that x squared equals 4x minus x squared okay we can

just subtract x squared on both sides and we get that zero is equal to 4x minus QX squared we can factor out an X

here actually we can factor out a 2x here to get that zero equals 2x times a

two minus X okay so obviously they intersect at x equals zero that's one of our endpoints okay and the other is two minus x equals

zero which is just going to be x equals two so that's going to be our other bound is x equals two so let's get on with it here first we immerse

circumference or circumference is 2 pi times R okay what's going to be our radius okay we'll think of this this is a is a cylinder when we rotate it around

this line x equals four our radius is going to be from our solid to our line

of rotation okay and well we know that if we want to find the distance of that line our radius we can say that the whole distance from Y from x equals zero

to x equals four is four okay and then we can subtract this distance from the y-axis to that exact same point and that's just going to be X right that's

just going to be X okay that's the distance that you've went over on the x-axis so right here you're gonna write four minus X next we need to figure out

the height okay so just the height of each of our shells okay now we're talking about this distance from our top curve to our bottom curve so what do we

do well we subtract the bottom curve from the top curve our top curve is 4x minus x squared okay so we'll have a 4x minus x squared here

and that's going to be minus our bottom curve which is x squared so we're gonna have an x squared here okay and then we have to do our width our width is just

DX or dy whatever variable we're integrating with and that is of course gonna be an X okay so now let's simplify this down a little bit so we can

integrate we get the integral from zero to two we can actually pull that 2pi out so why don't we go ahead and do that

okay and now we have just a four minus x times a four X minus two x squared DX now distributing here we're gonna get

that this is the volume of 2 pi the integral from 0 to 2 of this will now be

a 16x minus a 4 x squared minus 8x squared plus a 2 X cubed okay and of course well with the X on the end

but we can simplify these like terms here to get minus 12x squared plus 2x

cubed and we'll have DX okay so now we can just integrate and when we integrate here we get that the volume is equal to

2 pi times this will be just doing this really quick here for times sake 8x minus 4x cubed plus X to the 4th over

2 and that will be evaluated from 0 to 2 so doing our evaluation step here we get that 0 will kill everything off so let's

just worry about 2 we get 2 pi times 8 times 2 is 16 4 times 2 cubed 2 cubed is 8 5 times 4 is 32

Oh actually this is 1/8 x squared I apologize 8 x squared so we'll actually get a 32 here okay great so now we have a plus X

to the fourth over 2 X to the fourth ok 2 to the fourth and it's going to be 16 divided by 2 u is going to be 8 okay so we get that this is all equal to 16 pi

all right so question number four what is the average value of f of X equals cosine to the fourth of X times sine X on the interval from zero to PI so the first

thing you need to know before you start off here is that the average value of a function is equal to one over B minus a okay so in this case it would be PI minus zero okay

a is your the first endpoint and B is the second and that's the integral from A to B of f of X DX okay so let's just do this here we did that the average

value is going to be equal to 1 over like I said PI minus zero so it'll be 1 over PI and that'll be the integral from

0 to PI of f of X that's going to be cosine to the fourth of x times sine X

DX okay this is just going to be a simple u substitution here we'll set u

equal to cosine of X and that means that D u will equal sine X DX we also need to

worry about our endpoints here we have x equals 0 and x equals pi when x equals 0 you will equal what little cosine of 0

okay we just plug that in here cosine of 0 is 1 so our U is 1 and x equals PI that's our other endpoint okay and well

that'll make cosine of PI is negative 1 so that gives us u equals negative 1 okay and who seem to have missed a

negative here negative sine of X DX man all right so now we get is equal to since we have a negative sine X here this will be a negative 1 over PI and

this is the integral now from 1 to negative point of U to the fourth do you okay

now since our bounds are going from it's 1 to negative 1 and it looks a little bit better if we switch it back and what we have to do then is this integral so if we negate it this negative goes away

on the outside we is equal to 1 over pi the integral from negative 1 to 1 of U to the 4th to you

okay and moving further this is equal to 1 over pi times U to the fifth over v

over five evaluated from negative 1 to 1

ok so you can now say that this is 1

over pi times we'll just have a plugging

in 1 first 1/5 minus a be negative 1/5

ok which is just a 1/5 plus 1/5 and

that's equal to 2 over 5 pi ok so there you go that's your average value now any quick note here I wouldn't assume

symmetry unless you're absolutely certain there is symmetry okay so we're we're integrating from negative 1 to 1 ok with you to the fourth we know that there's symmetry ok don't worry about

what there is with x's ok then we had different bounds ok we're integrating from negative 1 to 1 of U to the 4th to you we know their symmetry so we could have just said that this was the same

thing as the integral of Q over pi times the integral from 0 to 1 of U to the 4th to you and you can do that if you're absolutely certain that there is symmetry

I've ran into cases in calc 2 where I wasn't sure if there was symmetry ok and and I always just made sure that I didn't go with it ok I was just like ok let's leave it as it is I'm not going to

change it to something that I don't know if there is a symmetry or not ok so just a little tip there now our next question is we want to integrate from zero to 1 e to the x

times sine X DX okay you can see here I kind of already spoiled it by putting it over here but we're gonna have to do integration by parts here and we have two functions being multiplied together and that's literally screaming

integration by parts okay there's no any any calc 1 integral that we can do here we're gonna have to do some integration by parts ok now we have this formula right up here we have to choose a U and

we have to choose a DV okay that's how integration by parts works so let's say that there's nothing really clear that we can see here of what will make one eliminate the other and you're gonna see

that's not even going to be the case we can't eliminate sine of X we can't eliminate e to the X so let's set u

equal to let's set u equal to sine of X ok and D you will now be cosine of X DX

okay well let DV equal e to the X DX okay and V equal e to the X okay so

let's see what happens now well we have the dist integral is gonna be equal to u times V okay u times V is e to the X sine X and only be evaluated on the same

bounds okay you don't need to change your bounds at all okay this isn't a u substitution okay this should just kind of and I know we're setting u equal to something but with integration by parts it's different okay you keep your

balance okay we're not we're not changing anything we're still integrating with with respect to X now this is minus the integral from 0 to 1

of VDU and VD U is e to the X cosine X and we'll have a DX there and you see the we're left with pretty much the same exact thing we have to do integration by

parts again so let's get on with it let's say u equal to cosine X okay and D

you will now be negative sine of X DX if we set u equal to cosine X then DV needs to be e to the X DX basically what's

left of that integral so that would mean that our V okay we have to integrate here and we get e to

the X all right so now writing this out we get to this is equal to still leave the x times sine X evaluated from zero

to one okay minus u times V so now our U is cosine X and we have e to the X so e to the X cosine X evaluated from zero to

one minus the integral from zero to one of VDU and that's a negative sine X DX okay times and of course times e to the

X so e to the X sine X DX now you might be saying here oh my god you know we're stuck again we have e to the X sine X

but that's good okay is if we have this integral back well remember what we started off with of course we had that the integral from 0 to 1 of 8 e to the X sine X DX

okay and what that means we can add that on both sides if we add this integral over ok yeah I just had a girl over then what we now get is that we have to

the integral from 0 to 1 of e to the X sine X DX is going to be equal to e to

the X sine X evaluated from 0 to 1 minus e to the X cosine X evaluated from 0 to

1 okay so now this is just evaluating but remember we're solving for the integral of e to the X sine X DX not to the integral of e to the X sine X DX so

we need to do is divide by 2 here okay and just for the sake of not writing this whole thing out again I'm just gonna erase this piece right here and

yeah okay so let's do that evaluation step we get that our integral is equal to e to the x times sine X the zero that will kill everything off because it'll

make the sign of well sine of zero is zero okay so we just have to worry about plugging in 1 and we get e times sine of

1 okay and then next we'll have minus we'll have what was your own kill everything off here no because cosine of

0 is 1 and e to the 0 is 1 so we need to do both terms we'll need to plug in 1 and then plug in 0 so plugging in 1 we

get e times cosine of 1 okay and then we'll have minus well like we said this either the zero is 1 cosine of 0 is 1 so

this will just be a 1 here okay so and of course this is over 2 so we'll just multiply this by 1/2 so we get that this

is finally equal to this will be a 1/2 times e sine of 1 minus e cosine of 1

plus 1 ok it's plus 1 now because we had to distribute that negative 1 okay so yeah

also one thing here before I erase the rest okay when you are integrating by parts the second time for something like this where you're not going to eliminate the e to the X you're not going to

eliminate the sine of X okay when you something like that you can't just just now pick your U as e to the X if you pick your U as e to the

X instead of that trig function that you are picking it as then everything kind of goes south from there okay so you don't want to do that so next we have the integral of sine - v

of x times cosine squared X DX okay now this is a trigonometric integral okay this is not going to be a simple u substitution at least not at first okay because if we set u equal to sine

of X well then D U is cosine of X DX and we're left with that cosine of that extra cosine of X there so what we have to do is look at okay how can we end up

doing a u substitution here okay well we have an odd power on our sine so what we can do is say that this is the

integral of sine to the fourth of x times cosine squared X times sine X DX

okay and now you have if you say are you equal to cosine X okay say are you equal to cosine X Y or do you is that this will get cancelled out with your do you well there's still the issue that we

have this sine to the fourth of X here so what we do is we use our trig identity that cosine squared plus sine squared equals one so sine to the fourth of X is going to equal one minus cosine

squared of X and that's going to be squared okay and then we're going to

multiply by cosine squared X and we'll have sine X DX okay so now we can do our

U substitution with what I just said u is gonna equal cosine X and D u will equal negative sine of X DX

alright so of course we don't have that that negative despair out here so what's gonna happen is we're gonna have

negative the integral of this will now be a 1 minus u squared quantity squared times u squared and then we'll have a D

you here okay so doing a little bit of foiling and then we'll do a little bit of distributing in the next step but we'll have a 1 minus 2 u squared plus u

to the fourth and that will be times u square do you okay we can distribute you that

you squared and we get au squared minus 2-u to the fourth plus U to the sixth do

you all right and we can integrate to get a negative u cubed over 3 minus 2-u

to the fifth over five plus U to the seventh over seven and that will be evaluated well actually we're not evaluating this one we're not evaluating this one because this is an indefinite

integral okay so now we can distribute through there we can multiply this negative through okay and we can get our use back to being in terms of X which is

going to be u equals cosine X that's what we substitute in so this is going to equal negative cosine cubed of x over

three plus two cosine to the fifth of x over five minus a cosine to the seventh

of X over seven and then we're gonna add C there you go that is your answer okay so we you know we looked at this

integral we saw that it's not going to be an easy u substitution but we needed to manipulate this to make it au substitution okay we saw that the sine had the odd power and what that meant

was we could pick off a sine and have that be canceled out with our D u okay all that was left to do at that point was to get this sine to the fourth of X

in terms of cosine rather than sine so we could have are you completely replace everything else other than that sine X DX

you know next what we have here is the integral from actor it's gonna be the integral of x over the square root of x

squared plus 4 times DX ok now for this integral right here you're probably thinking alright this is going to be an inverse trig substitution well you can

do that and I definitely did that before on a calc 2 test and it took so much longer than ahead to because this is just a simple u substitution okay just

you so so and so trust me I'll always be looking for that now since I'm that stuff like that ok so I don't want you to have to do that so please just check for Au substitution first if you set u

equal to x squared over x squared plus 4 here you get that D u is equal to 2x DX and you can divide by 2 on both sides to

get that D u over 2 equals x DX okay you know so so definitely beware of a calc 1 integral hiding in a calc 2 test now

what we can do here I guess we'll keep going here that X DX gets canceled out with the D u over 2 so we'll have an a 1/2 out here and we'll have a 1 over R

at x squared plus 4 is just gonna be rad you and we can rewrite that since it's 1 over R and you write that as U to the negative 1/2 of course we'll have that D

u here we already took care of that over 2 by putting outside the integral so integrating here we're gonna get 1/2

times this will be just a well we raise 1/2 by 1 which gives us 1/2 we flip it I multiply it so we get 2 here times U to

the one-half and that'll be plus C because it's an indefinite integral so this goes away and you're left with red

x squared plus 4 plus C okay so again I'll say the skin because like seriously this will save you so much time on a test if you just check for a

u-substitution before you do something like an inverse trig substitution which will most likely take you around five to ten minutes now we have our next problem up on the

board here the integral of 1 over x squared times the square root of x squared plus 4 DX now this I mean you can see that we're not going to be able to do your substitution here okay this

isn't an integration by parts it's not just a trig integral where we have to manipulate a little bit and then do it use of this is indeed an inverse trig

substitution so what's the substitution well this is since we have our we we've a plus here K rather than a minus we

know that this is a a tangent theta substitution ok now so we have X equal to a tangent theta what's a well a is the square root of

that number that's attached which is 2 so we have x equals 2 tangent theta okay

if x equals 2 tangent theta then DX is going to equal to secant theta d-theta

sorry 2 secant squared theta D theta ok so now we have our X in our DX so we could just go substitute this through right now the reason why I'm not going

to do that is because well you know we have an x squared plus 4 here and we have an x squared here okay so I'm not gonna have I'm not gonna rewrite this integral a million times and do step by

step by step what I'm gonna do is just solve for square root of x squared plus 4 and x squared right now okay so x

squared that's gonna equal for tangent squared theta okay x squared plus 4 is equal to 4 tangent

theta sorry for Kenya yeah for tangent squared of theta plus 4 okay and you can

factor out that 4 to get a 4 times tangent squared theta plus 1 which is

equal to a for secant squared theta now we have a square root of x squared plus 4 and that means we're taking the square

root of this which is just going to be hu secant theta so nice - what it mean we not only we don't have to rewrite

this integral a million times okay we can substitute in this x squared this rad x squared plus four and we can substitute in this DX and that takes

care of every single piece that was part of that integral okay so doing that we get that this is now the integral oh

well DX is secant squared theta D theta so we'll have a 2 secant squared theta here and we'll put a D theta out here

this is gonna be over x squared is 4 tangent squared theta and this is going

to be x AQ secant theta okay so the secant that this squared here will get cancelled out with the secant here so and this Q will get cancelled with this to youhere

so we can pull this 1/4 out and we have 1/4 the integral of secant theta over a naught 4 we already pulled the 4 out we

have a tangent squared theta and a D theta out here okay now looking at this this is pretty nasty we can't do au substitution I originally looked at this

and I was like oh we can just do a u-substitution but we have the squared on the wrong trig function if the squared was right here and then we could say U is equal to tangent theta D U is

equal to secant squared theta D theta we don't have that so whatever I do when I see you know what some trig functions that it looks really nasty and I'm not gonna I don't really know what to do

with it I break it down into sines and cosines if you break it down into sines and cosines well secant theta that's 1 over cosine

theta and tangent squared theta that's sine squared theta over cosine squared theta and we'll have a D theta

here and this cosine theta gets crossed out with this square here on the bottom cosine and you're left with 1/4 the integral

from this cosine will get flipped up it'll be cosine theta over sine squared theta and you'll have a D theta here okay and now you can do that u

substitution so we can do is set u equal to sine X or sine theta rather do u will be equal to now cosine theta D theta

okay and moving on up you are gonna get 1/4 the integral of U to the negative 2

D u ok or 1 over u squared ok so integrating that you are going to get 1/4 times a negative U to the negative 1

and that will be we're gonna have a plus C here but we can write this as a negative 1 over u ok we're gonna do that

and now we need to get this back into terms of X okay because that's what we started off with okay so we're gonna have to do that triangle thingy and all

that good stuff so let's start off here now we know that U is equal to sine theta so we can say that this is negative 1/4 times 1 over sine theta

plus C here but we need to get sine theta okay in terms of of X we need to get that theta in terms of X so what

we're gonna do set up that triangle okay the good old triangle we know that there will put our angle here and well right

here this is telling us that x equals 2 tangent data okay if x if x equals 2 cos theta then that is the same thing as

saying x over 2 equals tangent theta we know that tangent theta is equal to opposite over adjacent okay this is our opposite side this is our adjacent side

so we can right over we can write a 2 right here okay now if this is 2 and this is X and this is a right triangle

what is the hypotenuse well the hypotenuse is just going to be the square root of x squared plus 4 okay

you're doing a little bit of Pythagorean theorem there yeah so now what do we do well we have a sine theta here okay and

if we express sine theta with this triangle using that triangle we can get this in terms of X so sine theta we know

that sine theta is equal to well sohcahtoa opposite over hypotenuse which is x over the square root of x squared

plus 4 so there you go now we can get this thing into terms of X with this conversion here and we get negative 1/4

times this will be a 1 over x over R at x squared plus 4 but we can also write this as well this is 1 over we can say

rad x squared plus 4 over X and then we can write plus C all right so you know

right there that is your answer so you know make sure that you don't leave it I've seen I mean I've done this myself leave it in terms of you rather than convert it back to theta and then 2x

okay you need to make sure that you do that or that's going to be some pretty big points off all right so yeah all right so for our next problem we

have the integral from 0 to 1 of x squared plus X plus 1 over X plus 1 quantity squared times X plus 2 DX now

this is not something that we can do an easy u substitution with it's not an integration by parts it's not a trig integral or inverse trig substitution

this my friends is partial fraction decomposition so the first thing we need to do is write out our partial fractions ok

we ran out our partial fractions here we're gonna get a over X plus 1 this is squared so the next term is going to be

B over X plus 1 quantity squared plus C over X plus 2 and that will be equal to

this x squared plus X plus 1 over X plus 1 squared times X plus 2 now what do we do next well if we get some common

denominators here okay for all these then we can just set the numerators equal to each other okay we can figure out what that a and B and C are so here

if we get common denominators what is the a fraction missing it's missing a X plus 1 and it's missing an X plus 2 what

about the B the B is missing an X plus 2 and the C is missing an X plus 1 squared okay and we'll set that now equal to x

squared plus X plus 1 and we don't need to worry about the denominators anymore because now all of them are the same ok so doing a little bit of distribution

here and a little bit of all that good stuff we get that this is equal to x squared okay plus we have a 2x plus 1x which is

going to be a 3x and then we have plus 2 we have a B X plus well of actually we can write this just this B times X plus

2 for now we'll leave it and then we have AC times x squared plus 2x plus 1 and that's going to be equal to x squared

plus X plus 1 okay so now what we're gonna do is we're going to group these we're gonna we're going to basically combine like terms here okay now we'll have an x squared term on this side and

that will just be we have an a here and to see here and that are both being multiplied by a singular x squared so we'll have an a plus C times x squared

okay so that's our x squared term now we need to do our X term well here see we have the three a we have a B and we have

a 2 C so we're gonna have plus 3 a plus B plus 2 C and that will be multiplied for Phi X that we need to do just our

constant and that's going to be Q a okay so plus a 2 A plus 2 B plus C and that's

going to be equal to x squared plus X plus 1 okay now the coefficient on our x squared term here needs to equal the coefficient

on our x squared term here that means that a plus C is going to equal 1 okay

and that also means that 3 a plus B plus 2 C is going to have to equal 1 as well

and Q a plus 2b plus C needs to equal 1 because all of our coefficients here and our constant is both all of it is 1 so

that's why everything is equal to 1 so now what we can do here since we already have a two variable equation here we can use these two

equations and cancel out the B's to formulate another equation with just days and C's okay so let's do that if we

take this equation and multiply by 2 okay we're going to get that we'll get a

6 a plus

2b plus 4c is going to equal 2 and this

equation is Q a plus Q B plus C and that's going to equal 1 so subtracting

these two equations ok we get 4a the two B's go away and you have a plus 3c and

that's gonna equal just one okay bring this other equation down you have an A plus C equals one but we need to cancel

one of these out okay and let's get rid of the a alright so what we're gonna do here is we're gonna bring this down and we're going to multiply it if we want to cancel out the a we're gonna multiply it

by four so let me draw a better arrow

here sure x4 so we get a 4a plus 4c and that's going to be equal to 4 so now we

can subtract these and the forays go away you're left with a negative C equals a negative 3 which means that C

equals 3 if C equals 3 that means that a would equal negative 2 if a equals

negative 2 and C equals 3 that we can easily plug into one of these equations here to figure out what B will be okay so let's do that let's plug into our bottom equation here we'll get that 2

times negative 2 that's negative 4 plus 2 times a well our B we don't know that and our C is plus 3 get that equal to 1

so negative 4 plus 3 okay that is negative 1 we add that one to the other side we get the to be equals

to divide 2 on both sides we get that B is equal to 1 okay and now we will be plugging those in up top here and integrating to

actually get our answer so remember our a was negative two so now this is going to be a negative to

our B was a 1 and our C was a 3 so now we're going to integrate this okay and what we can do is since these are all separate fractions we can integrate them

separately okay so we'll have the integral all right over here the integral from 0 to 1 of

negative 2 over x squared plus 1 DX okay which is the same thing as saying put the negative 2 out here and 1 over X

plus 1 okay we'll have a plus the integral from 0 to 1 of 1 over x squared

plus 1 DX and then we'll have plus the integral from 0 to 1 we can pull that 3 out so we get 3 you need to go from 0 to

1 of 1 over X plus 2 DX ok now this this any girl is going to be easy the Santa girl is gonna be easy this integral is going to require a u

substitution okay you'll generally not get the nastiest integrals when you break it down like this mostly you'll be dealing with probably just like a natural log or you might have to do au

substitution very rarely do I see anything a little more challenging so you get u equal to X plus 1 here D u

will be equal to DX and you need to change your endpoints so what you'll get

is now if x equals 0 U is going to equal 1 and if x equals 1 then u equals 2 okay so now we can just kind of do these

integrals and then write this integral again we have negative two this will be x natural log of absolute value of x plus one evaluated from zero to one

we'll have I'll write this integral next plus three natural log of X plus 2 evaluated from zero to one okay which means that we can just erase

this evaluation and evaluate this whole thing from zero to one okay we can do that and then we're gonna add this integral here the integral from now one

to two of this twos a negative bar sorry it's going to be a one over u squared which is a U to the negative 2 D you

okay so this is now we'll keep this as negative two times natural log of X plus

one plus three times natural log of X plus 2 evaluated from zero to one for now and we will make this well this is

now a negative U to the negative one or in you can write this as negative one over U and that will be evaluated from 1 to 2 okay so now let's do other

evaluations all together here and well zero will that kill everything off here no it won't because you'll have a natural log of 1 that'll go away but you'll have a natural log of Q here to

deal with so zero will not get everything off 1 let's plug in 1 first we get a negative 2

natural log of 1 plus 1 which is 2 plus a 3 times natural log of 1 plus 2 is 3 okay and this will be minus remember we

said this term goes away because you have a zero plus 1 natural log of 1 is 0 but you do have this 3 natural log of 2

okay now we have a plus this is a negative one over two minus a negative one okay when you plug that in and

you're going to add these because it's two negatives and you just get a 1/2 here so simplifying you've a negative 2 natural log of 2 minus a 3 natural log of 2 which would give you a negative 5

natural log of 2 you have a plus 3 natural log of 3 and as we said here you get a 1/2 out of this there you go

there's your answer I believe this was one of the longest problems if not the longest problem that we'll be doing in this video so don't worry problems like this are over

so next we want to see is the function convergent or divergent we have f of X equal to the integral from 1 to infinity of x over X cubed plus 1 DX ok so we

want to see if this integral is going to converge or diverge now is this an integral that we're going to easily be able to do I mean we know that since we have this infinity here we'll have to

have a limit as T approaches infinity ok but here's the idea I mean this integral is going to be tough ok the center girl I don't even think will be able to do it so what we're gonna have to do is look

to the comparison theorem ok we can compare X over X cubed plus 1/2 let's get rid of that constant there let's get

rid of that plus 1 let's compare it to x over X cubed ok because that is equal to 1 over x squared and that we can do pretty easily that that's a integral

that we can do pretty easily so what's the relationship between these two since you're adding one here this denominator is going to be larger so the fraction will be smaller than this fraction here

ok so and of course we should know this by now this integral will be convergent so

since this is greater than this this will be convergent as well but I'll show you that real quick here so we have let's do the over here the limit as T

approaches infinity the integral from 1 to t of 1 over x squared DX okay because we have we should we should show now

remember we have to write that limit ok we cannot have an infinity as a bound okay that's something that math people hate so we have a limit as T approaches infinity and we just plug in T which is

basically still the same thing as integrating from 1 to infinity ok so now we can just do this integral we have the limit as T approaches infinity of

integrating this we get negative 1 over X ok we've done the integral enough in this video negative 1 over X and we're gonna evaluate that from 1 to T

okay that gives us the limit as T approaches infinity of negative 1 over t plus 1 now as T approaches infinity this

fraction right here it's going to go to 0 and we're just gonna have that plus 1 so this equals 1 now that since this is

a finite answer this integral converges ok and what that means is that by the comparison theorem this integral will converge as well

so here is an example of a of a good written out answer here we have the integral from 1 to infinity of 1 over x squared DX converges and it's greater than the integral that we've been

working with on the interval from 1 to infinity therefore the function is convergent by the comparison theorem okay that's all you really got to set now and function you can replace that

with this integral if you want okay it doesn't really matter so because it's asking if the function is convergent or divergent so yep so for our next question here we want to

find the arc length of y equals one plus six x to the three-halves on the interval from zero to one okay so here we have our arc length formula okay and

this is what we need to use for this problem okay so I'd remember this is the arc length formula not for parametric not for polar but for Cartesian okay

just for just for regular Cartesian no parameter and we're not talking about polar either okay you know they have each have their different formulas now

we use this formula okay we know we're integrating from zero to one and that's obvious so we can write our arc length is equal to the integral from zero to

one but what's going to go in this radical we have a dy/dx squared we need to find dy/dx alright dy/dx is gonna equal bring this

three-halves down we're going to get this is 18 over 2 this is 9 X to the 1/2

okay if you square that okay if you

square that you get 81 X okay so we're gonna put in 81 X right here all right

and this just becomes a pretty easy you substitution from here okay so we get U

is equal to 1 plus 81 X and D U is going to equal 81 DX okay we don't have an 81 just laying around here so we're going

to divide by 81 on both sides okay and also we need to change our bounds so when x equals 0 what will u equal well U

is going to be 1 plus 81 times 0 which is just going to be 1 and X when x equals 1 u equals 1 plus 81 times 1

which is 82 so we're now integrating from 1 to 82 okay so we have the integral from 1 to 82 of rad you

times D you over 81 so we can write it 1 over 81 out here and a D you on the inside ok so going through with this

further okay this is a U to the one-half so when we integrate this is 1 over 81 times 2/3 times u to the three-halves evaluated

from 1 to 82 ok Annan's well we have a 2 now 81 times

3 is 243 and we have au to the three-halves which means we'll have if you have anything to the three-halves what that means is that it's going to be

like if it was u to the three-halves it's you read you okay so for here it'll be 82 around 82 minus a 1 red 1 which is

just 1 and right there that is your arc length now what if we wanted to figure out the arc length function starting at this x equals 0 okay well if we wanted

to find the arc length function then all we do is we don't have any a top bound anymore okay or we don't have a number

and our top bound okay our top bound will become an X okay and it won't be an X here it'll be next year and that will change when we do our u substitution so

instead of an 82 okay you'll be starting out with X equal to X and that means you will equal 1 plus 81 X and that will be

what we substitute in as our bound right here so we all have 1 plus 81 X okay and then our bound here will be a 1 plus 81

X okay and then what we will end up getting okay is instead of an easy to

read 82 we'll have a 1 plus 81 X to the three halves minus one okay and that will be your arc length function denoted

s of X okay so that's all you do for an arc length function you just put in X in your top bound and you go for okay and that's going to give you the

the arc length at any x value okay assuming that you're starting at some X or starting at zero here so for our next question we want to find

the surface area of the solid obtained by rotating y equals the square root of nine minus x squared from zero to two about the x-axis now if we're trying to find the surface area of this salad we

need to use this equation the integral from A to B of 2 pi Y times the square root of 1 plus dy DX squared DX now I have this y circle because this is

dependent on the axis that we are rotating about if we're rotating about the y axis this becomes an X okay you have a DX over dy squared and you have a

dy ok so that's what changes so but we're rotating about the x axis here so that's what we're gonna be working with

okay so we'll just write a Y here now starting off here we need to figure out what our dy/dx is going to be okay so we

need to find that so right over here dy/dx is equal to what we have a 9 minus x squared to the one-half as our Y so we

take the derivative we'll have a 1/2 times 9 minus x squared so negative 1/2 it'll be times the derivative of the

inside which is negative 2x okay and all together that's equal to the twos go away and you're left with a negative x over the square root of 9 minus x

squared okay now we need to square this ok we need to square this because we have dy over DX squared on the inside so let's Square this let's get rid of this equal sign

because now it's not going to be equal anymore and we get that this is x

squared over 9 minus x squared ok and that right here is what we're going to be plugging in for our dy over DX

squared ok so let's do that we get that our surface area okay our surface area is going to be equal to the integral from 0 to 2 hey those are our

bounds and that's gonna be of 2pi times y which we're going to represent in terms of X because we're integrating with respect

to X maybe square root of 9 minus x squared then we have the square root of

1 plus dy DX squared which is x squared over 9 minus x squared DX okay so now we

can take out that 2pi outside the integral just to make this look a little prettier and we have that 9 minus x squared inside that radical still but

for our next radical well this is the same thing as if we get common denominators here we get a 9 minus x squared plus x squared over a 9 minus x

squared okay this 2 minus x squared and the plus x squared go away and you're left with a rad 9 over 9x 9

sorry rad 9 over 9 minus x squared ok and what you can do is you can just kind of get this radical out of here and just

put it on the denominator and make this 9 of 3 okay just doing all these steps at once by it like erasing and all that stuff here because I don't really want

to write this out like 5 more times ok I just four times take again so we get rid of these 9 minus X Squared's okay they cancel out so we get that our surface

area is equal to 2 pi the integral from zero to two of just three which is just going to be times three X evaluated from

zero to two okay and well the zero kills that off so we only need to worry about plugging into the d2 which gives us 2 pi times 6 which is 12 5

all right so as long as you know which formula to use okay if you're rotating about the x axis you're using the one with y if you're rotating about the y axis then your bounds are with respect

to Y okay you have a dy here and X here and you have DX over dy okay so yeah just just that really is the only hard

part about what we just did I mean to get this to simplify down usually how it ends up working at least in like half of the problems is that you have to get common denominators inside the radical

and then you get something nice where the this Y right here will cancel off with that dy over DX squared or something like that will happen

all right so next we want to graph the parametric equations x equals T squared and y equals T plus 1 okay now how do we

start off with this draw a key XY table okay T XY table is just the easiest way to go about this okay now it doesn't specify here it might specify under test

it most likely will but what I'm gonna do I just do this from negative 3 to positive 3 all right

and we will just figure out what our X's and Y's are gonna be so when T equals negative 3 what's your X well your X when you plug

in negative 3 for T you get 9 okay and that will be the same for positive 3 okay and of course you can just kind of fill the rest in when you have a negative 2

penny this will be for you know what before down here when it's one you'll get ones here in here what at zero you'll get a zero but what about with y

well this is just T plus one so you're gonna have negative 2 negative 1 0 1 2 3 4 and now you have your XY pairs okay so

you're inputting the parameter and this is your input column and then you're outputting an XY pair let's graph that

ok you are starting out with a 9 comma negative 2 so let's go out 9 spaces 1 2

3 4 5 6 7 8 9 okay you're going down to so first point is gonna be right here ok

now next really 4 comma negative 1 so 1 2 3 4 comma negative 1 next you've a 1 comma 0 so just right

here okay next you have a 0 comma 1 right here okay

give it 1 comma 2 okay that's going to be 1/2 next you have a 4 comma 3 okay so

this will be a 1 2 3 4 1 2 3 and lastly you have a nine comma four you're gonna go up another tick and you

have right up here there nine comma four okay so now you can trace this okay tracing this all hopefully I don't mess

up here got it and I get to that last dot there we go man that still looks pretty ugly all right well do a better job that I did that's

my that's my first recommendation okay I just have the inability to be able to draw I just made it worse and well here

this is T equals negative three right and this is T equals negative two so as T increases we're going this way okay so

if we need to give this graph in direction and it's of course going to be in the direction that T increases okay that's how you figure it out that's how you can figure out your direction well this was my T plus a negative 3 dot this

is my T equals negative 2 dot it's increasing this way so my arrows are gonna go around this way okay now next we want to find the second derivative of our set of parametric

equations here x equals T squared plus 1 and y equals T squared plus T okay now this is not something that you have to you don't have to memorize the formula

for your second derivative okay you don't what you need to know is that dy over DX of course I mean this isn't something okay this isn't something that

you have to know it's something good to know but you don't have to know it okay you can easily set this up because if you flip this you get dy over DT times

DT over DX your DTS cancel off and you're left with dy over DX so it makes sense okay it makes sense for this to happen okay and you can see more about

that in the actual video the explanation video that I did for derivatives with parametric but anyways what is going to

be our dy over DT we need to take a derivative with respect to t over here okay and that will be a 2t plus 1 next we need to take a derivative with

respect to T of our X equation and that will just be 2t okay and well that is

the same thing as 1 over 1 plus sorry 1 plus 1 over 2t ok we can split that up say that so that's your first derivative but we need to find the second

derivative and to find the second derivative hey remember you can't just square this and square this that's not

how it works ok what you need to do is wrong side again take the derivative

with respect to time of dy/dx and put that over DX DT ok now this also makes sense this is something I definitely

covered again in the explanation video for derivatives with parametric which is all these for all these problems I'm going to have a video that you could reference to I'll put that in the description kind of a little late to say

that cuz we're on problem 14 yeah so dy/dx now that's this piece so that's gonna go right here okay

so this is equal to the derivative with respect to time of 1 plus 1 over 2t and

then we're going to put that over DX DT which is 2t taking that derivative well really we only need to worry about the derivative of 1 over 2t because the

derivative of 1 is 0 so if we take this derivative here what we're gonna get is a little is a 1/2 times a this is T to

the negative 1 so this becomes a negative T to the negative 2 okay which is the same thing as negative 1 over T

squared and put that over 2t and then simplifying this down okay you get that this is equal to negative 1 over 4 T

cubed and right there is your second derivative okay so from that second derivative now you're able to see okay where's this gonna be concave up where's it's gonna be concave down and that may

be a question that you're asked okay that definitely would make sense for them to say okay where is this going to be concave up where is it's going to be concave down find the points of inflection something like that okay

okay so now we want to find the arc length of x equals cosine T y equals sine T from 0 to 2pi okay now for this question

here's your formula okay and you this is another formula you don't necessarily need to memorize okay as long as you know the arc length formula for just regular Cartesian no Parab at no no

parameter or anything like that you're fine okay because you can you can derive this and I did more about that in the

explanation video for this okay but you know this is probably it's probably easier to memorize it but you can always derive it if you forget this equation

okay so all hope is not lost now we're integrating from zero to 2pi here okay that's what this is saying and

DX DT well DX DT is that's going to be negative sine of T okay so so I'll actually write that right here DX DT is

going to be negative sine of T so dxdt squared is going to be sine squared T

okay and dy DT is going to be cosine of T so dy DT squared is going to be cosine

squared T okay so now you have what you need to put in to this formula already done so you don't need to write this formula a bunch of different times okay

so on the inside you have a sine squared of T plus a cosine squared of T and you have DT on the end this whole radical is

one that's our identity so really you just have the integral from 0 to PI to

PI of DT which equals T evaluated from 0 to 2 pi which equals 2 pi all right now

what is this I mean if you figure and the arc length of your unit circle because that's what the parametric equations of this are for okay this is your unit circle you

Circle is radius of one and we know that their circumference is 2 PI R well the radius is 1 so your circumference is

just 2 pi so we found this out with calculus now for our next question we want to graph our equals cosine of 2

theta okay so graphing polar now the first thing we want to do is view R and theta as they as if they were Cartesian coordinates okay so we're gonna have our

on our y-axis and theta on our x axis okay now what is our gonna rage between well it's gonna if it's gonna range

between 1 and negative 1 right cosine of 2 theta is not going to end up being any smaller than negative 1 and any larger than 1 so now we need to figure out

what's gonna happen with our theta axis here okay what are going to be our individual tick marks here well do you that well of course we know we're gonna start off at 1 right with cosine we

start off at 1 but when are we going to get to 0 and that's what our little interval between tick marks is going to

be so we need to figure out when does cosine of anything equal 0 and that's

well the the soonest is when you get PI over 2 okay so you want to theta equal PI over 2 and if you divide by 2 on each

side you get theta equals PI over 4 so that's going to be your next tick mark all right so here we're gonna write PI

over 4 and then PI over 2 and 3 PI over 4 PI and we can keep going a little bit

here let's go to 2 pi here we can write 5 PI over 4 and then this will be 3 PI

over 2 and then we have 7 PI over 4 and 2 pi okay

so we start off at 1 we go down to PI over 4 we go over to PI over 2 up to 3 PI over

4 and that further up to PI and then we're just gonna repeat that cycle okay so now that we have our two theta

graphed as as Cartesian coordinates we can transfer that over to a polar graph all right and I know we were the polar

graph we just have this polar axis which is the the positive x-axis but I'm gonna kind of just use these two lines here it's kind of like guidelines I mean that

they just they really help so I'm gonna keep them in now we're starting at well

theta equals 0 and R equals 1 so we're starting right here right so we go from

as we go to our angle of PI over 4 which our angle PI over 4 is right here okay we're going to go to 0 okay so what

that's gonna look like is this okay so you can see that that we're starting off off really quickly decreasing and then

our third we're starting off decreasing pretty slowly and then we're increasing rapidly as we get to PI over 4 ok that's why this curve looks exactly like this now as we go to PI over 2

okay we're going in the negative direction out to negative 1 so what this ends up looking like is this okay next

we're going again we're a negative arc we're going from PI over 2 to 3 PI over 4 so what this looks like

is this ok and you can just just keep following this pattern honestly okay once you get the hang of what this graph

will look like you can just kind of keep going right and right there you have it okay that is R equals cosine of 2 theta

all right now for our next problem here we want to find the slope of the tangent line for R equals cosine of two theta again but at theta equals PI over four

okay and here's your equation here's your D now this is something okay if you don't remember it that's not a big deal okay this is something that we figured out in the explanation video for derivatives with

polar so again that'll be linked in the description if you want to check that out okay so now what we need to do is we have we

have four things to you to solve for here okay we have G our D theta we have

R we have cosine theta and we have sine theta okay now or we know that's cosine of two theta but plug in PI over four okay

when you plug in PI over four you get cosine of PI over 2 which is 0 for dr d-theta that's the derivative okay of

this so so that's negative 2 sine of Q theta when you plug in a PI over 4 for

your theta you get a bolt you're gonna get a PI over 2 right same as before and now you've sine of PI over 2 which is 1

1 times negative 2 is negative 2 cosine of theta cosine of PI over 4 is rad 2 over 2 and sine of PI over 4 is rad 2

over Q as well so now you have or everything that you need to plug in ok we now get the dy/dx is gonna be equal

to well we don't even need to worry about these are parts we don't need to worry about the second part here because we know that this orange is 0 so it's gonna can't fall off that wall term now

dr d-theta well you know that's negative 2 times sine of theta that's rad 2 over 2 that's gonna be over d rd theta which is negative 2 cosine of theta is rad 2

over 2 okay and obviously that is going to be equal to one which is your answer all right so really you know nothing nothing

too difficult right there as long as you know this formula okay and if you don't as I said before you can actually derive it now next we want to find the area

bounded by R equals cosine of theta from 0 to PI over 2 okay and when I said no this is your o

Kathina okay it's data is between 0 and PI over 2 they're rigged up messed up so what we need to know here is that when we

integrate with polar our area is equal to the integral from A to B of 1/2 R squared D theta okay that is what we

need to know before we start off ok if you know that then this problem becomes pretty easy ok this problem will become pretty easy you just have that your area

you're gonna integrate from 0 to PI over 2 and you have a 1/2 times what your are

your R is cosine theta squared is cosine squared theta and then you have D theta okay so you can use your half angle

identity simplify this down you can pull out a 1/2 that you get from the half angle identity and you have a 1/2 here that makes one-fourth and on the inside

you're left with a 1 plus cosine of 2 theta ok so use the half angle identity here to get 1 plus cosine 2 of 2 theta

over 2 and that over 2 I took out with this one out to get 1/4 on the outside

ok so now we have a 1/4 we can integrate this to get a theta plus sine of 2 theta

over Q and that's going to be evaluated from 0 to Q naught 2 pi PI over 2 and

now we can evaluate so we get 1/4 times pi over 2 plus the sine of 2 times pi

over 2 is just pi sine of pi 0 so we don't need to worry about anything there so we'll have pi over 2 minus the next term which will is going to be 0 plus sine of zero is again it's going to be

zero so really all we have is a PI over two here so we have 1/4 times 2 PI over 2 which is going to give us PI over 8

all right so again just really knowing before meal here and that's basically it all right so if you were to get any kind of problem with with sequences here this is kind of what I would expect you to

see okay it would be kind of testing your knowledge on just everything about sequences so is the sequence increasing or decreasing is it monotonic or not monotonic is it bounded is it convergent

or divergent basically it's just gonna ask you everything about this sequence and our sequence here is 1 over 2 n plus 5 so let's answer the first question is it increasing or decreasing well we can

check that I'm figuring out what a sub n plus 1 is okay a sub n plus 1 means we're just gonna plugging in n plus 1

for n if we do that we have 1 over a 2 times n plus 1 plus 5 okay distributing that 2 through if we get a

1 over 2n plus 2 plus 5 which is going to be a 1 over 2n plus 7 so now if we look at 2n plus 1 over 2n plus 5 in

comparison to 1 over 2n plus 7 we realize that this denominator is smaller so the fraction is bigger okay so a sub n is greater than a sub n plus 1 which

means that this is strictly decreasing and now we can move on to the next question is this monotonic or is it not

monotonic now what does that mean what does monotonic mean well sequence is said to be monotonic if it is strictly increasing or strictly decreasing okay

so something like sine okay we're just going back and forth and back and forth is not going to be monotonic because it's not strictly increasing or strictly decreasing it's doing both so here we

just have something that's strictly decreasing so it is monotonic next we're asked is it bounded it is the sequence

bounded well let's look at kind of what this would look like now it's gonna model something like 1 over n all right

you can basically say that it's gonna look something like 1 over n so what you'll have is something like this okay that may not be exactly what it

looks like okay but it's gonna be along those lines okay because you've something that looks like one over N so you can kind of see here that you're

going to have it's gonna be bounded on the bottom you can definitely see that but since it's strictly decreasing it's also going to be bounded on the top okay

so that means it is bounded now what is it bounded by well here of course it's bounded on the bottom by zero but where does this top out well it's actually going to be where this point is the

first point and the first point is where N equals one so where does it if we have our sequence here 1 over Q n plus 5 if

we plug in a 1 for n excuse me if we plug in a 1 for n we will actually get what that a sub n will be okay

so here we go a sub 1 equals 1 over Q times 1 is 2 plus 5 is 7 so a sub 1

equals 1/7 and that is your top bound so it's bounded by 1 7 and 0 so you can say

bounded so 0 is less than a sub n is less than or equal to 1/7 all right now lastly is this convergent or divergent

okay now normally you can take the limit as n approaches infinity okay you can take the lemons n approaches infinity of your a sub N and then see if it's going to converge to anything or it's going to

diverge and you still could do that here but here's something we do know it's it's monotonic okay it's monotonic and it's bounded which means that it's going

to be convergent by the monotonic sequence there okay so so just a quick recap of the monologue sequence theorem it says that

if you have a sequence that is monotonic and bounded it will be convergent and it's going to be convergent by the monotonic sequence there okay

all right so question 20 says given the F partial sum for the series these sum from N equals 1 to infinity of a sub n will the series be convergent or divergent and if it's convergent find

the sum ok so you're giving your F partial sum here now with sequences we graph a sub n

versus n but with series we graph S sub n versus that case we graph the art our partial sums so instead of taking the limit as n approaches infinity of a sub

n which we would do for sequences to see if it converges or diverges for series we take the limit as n approaches infinity of S sub N and s sub n is right

here that's that 6n squared plus 7 n plus 2 all over for n squared plus 2n minus 85 okay now if you want to do this

the hard way divide everything by N squared or use a little beat oh you don't have to do that though hey we don't need to worry about the 7 under the - ok we won't worry

about the highest power event we want to worry about the highest power event here and the coefficients on those we both we have an N squared on the top and bottom and the coefficients and is a 6 & a 4

simplifying that that's three halves and that's what the limit will equal so it's convergent okay so it's another thing

you can write it's convergent and the sum is the product of that limit which is three halves so for question 21 we want to see if the

series below is absolutely convergent conditionally convergent or divergent and our series is the sum from I equals 1 to infinity of 5n squared plus 7 n

plus 4 over 300 plus 2 n plus 1 and remember for 4 for our series for trying to see if they converge or diverge we should be using the acronym guy Clarke

okay now that's the for those of you that are new that's the acronym that I use it's kind of like my checklist for the series tests ok and I do it in this order so I check for the test for

divergence first the integral test the comparison test the limit comparison test alternating series ratio and then lastly the root test okay

so firstly well of course we have the first part of the the D okay that's the divergence test so let's try the divergence test for this we take the

limit as n approaches infinity over a sub n which is 5n squared plus 7 n plus

4 over 3 N squared plus 2 n plus 1 and of course just as we did the limit for the last example we're gonna do the

lemak here and and by looking at the N squared coefficient which is just 5 over 3 that will be the product of our limit and with the test for divergence set

what the test for divergence tells us is that if the product of our limit is not equal to 0 which it's not here then

these series is divergent by the test for divergence in my recommendation of course you know if you're following this checklist it

works out okay but definitely always try the test for divergence first unless you have something like a factorial or you know something that it's obviously going to be a ratio test or it's obviously

going to be some other test I would go straight to that test but really you know if it's a series that you're not too sure about always start off with the test for divergence

next we want to again see if the series below is absolutely convergent conditionally convergent or divergent and we have these sum from N equals 1 to

infinity of 2 over 6 and minus 1 now here let's do our series tests ok first off we have the test for divergence will we be able to do that here we can but

the limits can equal 0 so the test for divergence is not gonna help us what about the integral test well actually that seems like it could work here ok this is going to be something that's

easily able to be integrated ok we can write this as 2 the sum from N equals 1 to infinity of 1 over 6 and minus 1 and

that's something we can easily integrate ok so let's do that we're gonna compare this to the function f of X equals the

integral from 1 to infinity ok 1 to infinity N equals 1 to infinity of 1 over 6x minus 1 DX now to make sure that

we can do this what we need to do is make sure that this function is going to be positive continuous and decreasing along this interval from 1 to infinity

ok so first let's check if it's positive well of course every term here you can see that every term will be positive ok for every and whatever you plug in for X you're gonna get something positive ok

it's all positive so that's gonna work is it continuous well you're not gonna see there's no way you can get any kind of discontinuity here ok from 1 to

infinity that is so it's going to be continuous and was it decreasing what's actually going to be easier to see if this N equals 1 to infinity the the a sub N here if that's willing to be

decreasing you can look at the next term in the series by plugging in n plus 1 for N and when you do that you get 6

times n plus 1 minus 1 which is equal to 1 over 6 n plus 6 minus 1 which is 6 7 plus 5 you can see that this denominator

is much bigger than this one so this fraction will be greater than this one which means that it is indeed decreasing so now these pasture

conditions you can actually go about integrating this okay so if you're gonna integrate this I'll bring it over here you have to take the limit as T approaches infinity because you cannot

have an infinity as a bound you have to have a quantity that approaches infinity so you take the limit as T approaches

infinity of the integral of 1 to T of 1 over 6x minus 1 DX okay this is just a u-substitution

it set u equal to 6x minus 1 and D u equal to 6 DX divided by 6 on both sides and there you go

now you have to fix your bounds but in terms of u if x equals 1 then what will u equal well it's going to be 6 minus 1

which is 5 and when x equals T u doesn't matter if you're multiplying by 6 it doesn't matter that you're subtracting 1 U is going to equal the T okay the only

time it would matter is if it was something like u equals 1 over X okay which would be a weird substitution but if you had to do something like that u equals 1 over X well then you have 1

over T which is going to be 0 so that's the only time that something like that would matter so kind of just substituting in here we get the limit as

T approaches infinity the integral from 5 to T of this is going to be we're

gonna bring out this 1 over 6 and we have a D you over you okay which gives us the limit as T approaches infinity of

1/6 the natural log X to the value of u from 5 to T and evaluating that you get the limit as T approaches infinity of

1/6 times the natural log of T minus natural log of time and you can see that that is going to be divergent by the e odd by the test for

divergence okay we're sorry about by the test for divergence what do I think this is going to be divergent by the integral test and why it's gonna be divergent is because this integral right here will be

diverging okay when you that natural log of T as T approaches infinity we'll get an infinitely large number here okay so since that equals infinity since your integral diverges your series will

diverge as well that's what the integral test tells you so number 23 we're again asking absolute convergence conditional convergence or divergence for now the series from N

equals 2 to infinity of 2 over the natural log of n now here you use the test for divergence you'll get that it's equal to 0 so that fails ok the integral

test you're not gonna be able to do anything with that here that fails comparison test well possibly you're not gonna you can't really rule out that possibility and you can't really rule

out limit comparison test either alternating series no this isn't an alternating series and ratio Andrew I don't I mean root will definitely not work and ratio I don't really see how

that's gonna help here so that's gonna be a no now it's ok to leave question marks like that you know maybe comparison test will work but I'm gonna reserve that you know for later that's

gonna be like the last case scenario or the worst case scenario if maybe some of these other tests won't work a little

bit better ok so here let's try using a comparison okay because we all we already know that none of the other tests are gonna work maybe let me comparison but that's we need to check

comparison first ok well first we can rewrite this as two these sum from N equals one to infinity of 1 over natural

log of n ok that's something we can do that's gonna make the comparison a little bit clearer so what are we gonna

compare 1 over natural log of n 2 and I've done this before in in another video because it's extremely hard comparison to really kind of see I mean

it's not as intuitive that's why I say it's hard but you want to compare this to actually just 1 over N ok why do you

want to compare it to 1 over N well 1 over natural log of n is actually going to be greater than 1 over N because natural log of n is less than n ok for

any n for any on this interval at least and this should be an N equals 2 because we can't have a natural log of 1 ok so

what this tells us since we know that the sum from N equals 2 to infinity of 1 over n diverges ok that means that this series is going to diverge as well by

the comparison test so here's kind of a nice word Lao answer explaining kind of kind of the whole deal here you know so since the series these sum from N equals 2 to infinity of

1 over n diverges ok since it diverges and that 1 over N is less than 1 over natural log of n we know that the series

the sum from N equals 2 to infinity of 2 over natural log of n will diverge by the comparison test ok so that's kind of the idea if one of

red diverges so if it diverges and there's something greater than that of course it's going to diverge as well ok it never really has the chance to be convergent are you so for our next problem here

again absolute convergence conditional convergence or divergence but now for these stuff from N equals 2 to infinity of 1 over N cubed minus 1 ok now this you can use the test for divergence not

going to work hey the limits can equal zero you gotta move on integral test not gonna work comparison though maybe that'll work ok let's try it let's try a

comparison here let's compare 1 over natural log or sorry I'm selling the last problem 1 over n cubed minus 1 to 1 over n cubed

what's the comparison there well 1 over n cubed right this denominator and cubed is gonna be greater than n cubed minus 1

so 1 over n cubed is actually going to be less than 1 over n cubed minus 1 and that's a problem because we know that 1 over N cubed converges so 1 over n cubed

would have to be greater than 1 over m cube minus 1 and it's not so what we have to do is use the limit comparison

test ok so basically these are our series ok our series is the a sub n the

1 over N cubed minus 1 and the compared series is B sub N and what the limit comparison the limit comparison test is telling us is that we need to take the

limit as n approaches infinity of a sub n over B sub n ok that's the first step we'll worry about the other stuff later so we we do that we take the limit as n

approaches infinity and a sub n that's 1 over n cubed minus 1 so let's write that down and we're putting that over 1 over

n cubed and we can this look a little bit nicer we get the limit as n approaches infinity of M cubed over m

cube minus 1 and of course you can see that this limit is going to equal 1 ok the limits can equal 1 1 is greater than

0 which is one of the conditions of the letter comparison test okay it needs to be greater than zero and it needs to be finite

now if the product of your limit which ours is one if it passes both of those conditions which it does here then what that means is either both this series we're going to converge or both the

series are going to diverge okay what that means is that by the by the limit comparison test this series is going to converge and the reason for that is

because these series from N equals 2 to infinity of 1 over N cubed will converge okay now remember we're wondering about conditional convergence or absolute

convergence okay but you can see here that it doesn't matter if you're taking an absolute value here okay this is still gonna be the exact same thing okay you're always going to get positive terms so the absolute value does nothing

for you and what that means is that this is actually going to be absolutely convergent so you can say right convergent by the limit comparison test and then you can say I drew a line through this because

I'm so tired at this point okay absolutely convergent because it is convergent and all of your terms are positive okay so that's how you can kind of explain how it's absolutely

converging as well okay so for our next problem we again want to see if the series is going to be absolutely converging conditionally convergent or divergent inter-syrian now

is a sum from N equals 1 to infinity of negative 1 to the n minus 1 times N squared over n cubed plus 2 here we could do tests for divergent it's not

going to work integral test in comparison test limit comparison not gonna work here ok need any positive terms any way and we don't have positive

terms we have that negative 1 to the N minus 1 so we're gonna have to do an alternating series test let's work with that what the alternating series test tells us ok first we need to figure out

what our B sub n is going to be okay this is our a sub n when you define our B sub n and our B sub n is just going to be the a sub n but taking out whatever

makes the the a sub n alternate which is that negative 1 the N minus 1 so our B sub n is just going to be N squared over

n cubed plus 2 now what we have to do is see if this is decreasing and then if it is let's take the limit as n approaches infinity and Steve that's going to equal

0 if it is that we know this conversion by the alternating series test so first let's see if this is decreasing let's see what B sub n plus 1 is going

to be B sub n plus 1 is just going to be a n plus 1 quantity squared over n plus

1 quantity cubed plus 2 okay now we don't really need to distribute this the foil this out and do like a big

expansion here because you can kind of just like think about this logically ok this is obviously going to be smaller than this the reason why is because your

denominator is being raised to the 3rd power so if you add 1 to n okay and then you cube that it's gonna make a lot more difference here then it will here

because n plus 1 is being not just not cube it's being squared okay so there's gonna be a bigger difference on the bottom so your bottom is going to be bigger and that means that your old fraction

will be smaller okay so B sub n is greater than B sub n plus one okay so just think about that

logically now next we need to see the limit as n approaches infinity of B sub n Palumbo's n approaches infinity of N

squared over N cubed plus 2 now well that equals 0 well yes because the highest power of n is on the bottom and it's not on the top so it's going to

equal 0 all right so that's the other part ok the limit equals 0 so that is checkmark checkmark

it's convergent by the alternating series test now remember okay and I've done this before we're to having I've just figured out if it was convergent or divergent

and left it okay actually got points after that was really met but you want to see if it's absolutely convergent or conditionally convergent so what you

need to do is take the absolute value of your a sub F T okay so let's do that down here we have these sum from N

equals 1 to infinity of negative 1 to the N minus 1 N squared over N cubed plus 2 we can take the absolute value of it which means we're just gonna drop the

negative 1 to the N minus 1 and be left with these sum from N equals 1 to infinity of N squared over N cubed plus

2 okay so now what you need is another round of dry Clarke okay so it's basically like two problems in one now

looking at this you test for divergence the limit will equal 0 what about the integral test the integral test actually

might work here okay so we can compare this to the integral from 1 to infinity

of we can say that this is x squared over X cubed plus 2 now looking at this is it going to be

positive continuous and decreasing on the interval first off it's all the terms are positive so yeah it's going to be positive not going to be

discontinuous anywhere so it will be continuous and is it decreasing okay well it's actually going to be easier to see if it's decreasing from the ends

here we can look at N squared over N cubed plus 2 versus n plus 1 squared

over n plus 1 cubed plus 2 but we just did that ok we just did that up here so we already know that yes it's going to be decreasing and that passes our three

conditions so we can actually take this inner integral ok so let's do that we get the limit as T approaches infinity

because we can't have infinity as a bound integral from 1 to T oh we have an

x squared over X cubed plus 2 DX okay we need to use up if we need a u sub so are

you it's going to equal x cubed plus 2 or D u we're going to equal 3x squared DX and we're going to need to divide off

that 3 and we need to also do our bounds when x equals 1 you will equal 3 when x

equals T you will equal T okay so there's all that so now let's go back

over here we now have the limit as T approaches infinity the integral from 3

to T of well this is ad u over 3 now so we're going to plot 1/3 and we have a 1

over u so we'll just have 1 over u du u which is going to give us the limit as T approaches infinity of 1/3 natural log

absolute value of U from three to tea okay and if I don't really have too much room so I'm just gonna do this here

natural log of t minus natural log of three you can see that it's going to be divergent it gives you about natural log

of T and T is approaching infinity okay so since that's happening that means that this is conditionally convergent

because this is divergent by the integral test and I wouldn't have reavie at these as I T and ast and things like that on the test I would just write it

out so you don't get points off because you have like a strict teacher or something like that but yeah so this is

conditionally convergent why is it conditionally convergent and not absolutely convergent well that's because it was convergent just by the

alternating series test the regular series was but we take the absolute value it was divergent and that's what means conditionally convergent okay if it was converging here then it would be absolutely convergent

are you so next again we want to see if the series is going to be absolutely converging conditionally convergent or divergent and our series now is the sum from N equals one to infinity of

negative 1 to the n plus 1 times n to the 4th times 4 to the N over N factorial ok so you see the factorial there okay and what you need to be

immediately thinking is no I can't do any test but the ratio test okay it has to be the ratio test you cannot do anything else you need to go to the ratio test that's it

so the ratio test tells us that when you think the limit as n approaches infinity of the absolute value of a sub n plus 1

over a sub n ok so let's do that here we get the limit as n approaches infinity of the absolute value of now there's gonna be really messy at first we need

the a sub n plus 1 let's plug in let's plug in and then plus 1 wherever we see an N now before we do that this absolute value is just going to get rid of this

negative 1 anyway so really there's no sense of doing extra work let's just forget about that negative 1 so here we have a n plus 1 to the fourth power

times a 4 to the n plus 1 and that's going to be over a n plus 1 factorial

this will all be over n to the 4th times 4 to the N over N factorial ok now we can clean this up

we'll get the limit as n approaches infinity of the absolute value of n plus

1 to the fourth power over n to the fourth times a 4 to the n plus 1 over 4

to the N and that's going to be multiplied down by an n factorial over n

plus 1 factorial ok so the limit rules say we can take the limit of each of these things independently and also before we even do that you can kind

of see how it grouped each of these things that we took the a sub n plus one up and their third counterpart when we did the a sub n I group them together because you're gonna see a lot of really nice cancellations okay and it's really

nice to see that see that when you group it this way so we take the limit as n approaches infinity of well actually what are we gonna have here I don't

actually yeah we should do this okay this first piece right here you could actually just look at it and tell me the limit of it okay we even end to the

fourth over n plus 1 quantity to the fourth and when we do our expansion here you're gonna see that the what will the first term be it's gonna be n to the

fourth okay so nothing else matters just the end of the fourth and the coefficient on that is going to be one the coefficient on the bottom and to the fourth is 1 so really the limit of that

is just going to be one okay now for the n plus 1 over 4 to the N well when you divide this out you subtract the

exponents and you just get four and for this n factorial over n plus 1 factorial remember that n plus 1 factorial is the

same thing as n plus 1 times n factorial and the n factorials cancel out and you're left with 1 over n plus 1 so really you're set the limit as n

approaches infinity of the absolute value of 4 over n plus 1 now as n approaches infinity your denominator will get infinitely large and that means that this will go to 0 0 is less than 1

okay because remember what the ratio test we compare this to 1 if it's less than 1 it's absolutely convergent if it's equal to 1 it's it conclusive and if it's greater than 1 well then it's

going to be divergent but here it's less than 1 so that's absolutely convergent by the ratio test and one thing you know one little piece of advice

whatever you know like whenever we had a really really tough series or anything like that my ta at the time would say one in doubt ratio test and really I

that's pretty much what you should be doing I mean what in doubt ratio test if the ratio test doesn't work then you know maybe move on to something else but if you're faced with something really hard

chances are the ratio test will do some damage to it all right so here again proving up so the convergence conditional convergence

or divergence but now for the sum from N equals 1 to infinity of 3 n cubed plus 1 raised to the nth power over 6 and to the 4th minus 2 raised to the 8th power

and I'd here put a big checkmark next to the root test that's your go-to okay don't this is kind of like seeing a factorial on a series and going right to

ratio test you see everything to the unpowered okay because you could rewrite this by taking apart like this then putting everything to the nth power it's

the same thing okay when you see something like that root test that all day okay so when we do the root test all the root test is telling us to take the

limit as n approaches infinity of the nth root of the absolute value of a sub n so let's go ahead and do that here

take the limit as n approaches infinity hello the nth root of the absolute value of 3 n cubed plus 1 over 6 n to the 4th

minus 2 and that's all raised to the N forgot to put that in all right whatever and the nth power cancels out with the nth root and you're left with the limit

as n approaches infinity we don't need the absolute value signs here yeah that's not going to change anything you get 3 n cubed plus 1 over 6 n to the 4th

minus 2 now you know that this limit is going to equal 0 any advice power events on the bottom 0 is less than 1

and so this is absolutely convergent by ID root test okay the root test is pretty obvious when you need to use it okay it'll literally is only useful for

something like this so and if it fails if it's inconclusive and you have something like this test for divergence okay the test for divergence will most likely prove it to be divergent that's usually what's happened to me at least

so I definitely will pass that on to you and and say yeah go test for divergence are you so for this next problem here we want to see if the series is just a kind of urgent or divergent we've the sum

from N equals 1 to infinity of cosine of n over N squared now here this series is is a little interesting you're not going

to be able to prove it with any of anything with your with die Clarke okay because here's the idea if you start listing off terms for this series okay

you you get cosine of 1 and I'm just gonna talk about the numerator here

cosine of 2 cosine of 3 cosine of 4 and let's go cosine of 5 + cosine of 6 and

well cosine of 1 and you can look at this for yourself remember these are these are in radians okay so cosine of 1 that's in the first quadrant that's going to be positive

cosine of 2 that is going to be in the second quadrant and that's negative cosine of 3 that's in the second quadrant negative cosine of 4 that's in

the third quadrant that's gonna be negative cosine of 5 that should be and the should be in the fourth quadrant and that's going to be positive and cosine

of 6 is in the fourth quadrant that's positive so you see that this not the signs are not alternating okay and you know they're they're not they're all

positive or either all negative so what this means is that you're going to have to prove absolute convergence okay if you prove absolute convergence then you

can prove that this thing is convergent okay so we can what we can do here is take the absolute value of this okay let's take the absolute value and

actually so we don't wreck the original series I'm gonna do that down here now where does this absolute value actually apply well the N squared is going to

stay positive but this cosine of end is what's alternating between negative and positive well not alternating it's just moving

between negative and positive okay so what is this absolute value doing for us well that means that instead of being between negative 1 and 1 this cosine of

n it is just going to be between 0 & 1 okay so what that means is that we can

compare this to 1 over N squared and know that cosine of n the absolute value of it over N squared is going to be less

than or equal to 1 over N squared okay and what that means is that this series is going to be convergent by the

comparison test if it is convergent by the comparison test that means that we have absolute convergence and if we have

absolute convergence that means it is therefore convergent okay so there is an example of where absolute convergence really really comes into play okay so

it's not something just useless that you learn so last two problems here all right we have 29 find the Maclaurin series for F

of x equals x cubed times e to the negative 3x squared okay so now how do we even go about doing this well we don't need to start from scratch because

we know the Maclaurin series for e to the X okay and that's something that you should know if you don't have it memorized it's not the end of the world

you can start from scratch finding the X and then find x cubed times e to the negative 3x squared okay so it's not the end of the world if you can't memorize

it but this is X to the N over N factorial okay I would definitely make sure that these are memorized though because I mean listen if you have to

start from scratch you're taking a lot of time that maybe 10 extra minutes or 15 extra minutes you know depending on how fast you are started from scratch with Maclaurin series but you know it's

it's that's a lot of extra time to be taking it on a timed test so if we know that this is e to the X well then we can figure out what e to the negative three

x squared is we just plug in a negative 3x squared wherever we see an X and that's gonna give us negative 3x squared

to the N over N factorial which we can write to be a little bit nicer as they sum from N equals 0 to infinity of

negative 3 to the N over or not not over times X to the 2n over N factorial and you can even take this out further and

write it as negative 1 to the N times 3 to the N it's up to you now we can go just our last step here X cubed times e

to the negative 3 x squared and that is going to be equal to be sum from N equals 0 to infinity of now we're just going to multiply through that X cubed

ok and that can go inside the Sigma it's not dependent on n at all so it doesn't really matter whether we have an inside or outside but if we multiply here we add the exponents getting we get 2n plus

3 so we just have a negative three to the N times X to the 2n plus three and that's going to be over n factorial and

right there you have your Maclaurin series representing X cubed times e to the negative 3x squared okay so moving on to the last problem

Oh last problem here find the Taylor series for f of X equals 1 over x squared at a equals negative 1 now this is something that we're gonna solve by starting out from scratch okay this

doesn't match any of our templates relieved that glowed that all any of our common Maclaurin series that closely so what we're gonna have to do here is just start from scratch okay so in the case

that you do have just do have to start from scratch you don't remember one of the power series or sorry one of the Maclaurin series that are common you know here's how you kind of go about starting from

scratch so the first step is going to be to take derivatives of f of X and find the nth derivative of X ok so we have 1

over x squared here we're going to write that as X to the negative 2 okay now F

prime of X that's going to be a negative two times X to the negative 3 now to make this a little obvious I'm more

obvious that there's a negative 1 here okay or a negative here that's what I'm gonna do I'm gonna put a negative 1 out here okay just gonna make things a

little bit more obvious because remember we're looking for a pattern here and when I take the double derivative here what happens is well I this negative 3

come down and cancel out this negative 1 so right now I have basically a 3 times 2 times 1 times X to the negative fourth

ok when I take the third derivative I end up with a what I have here a

negative 1 times 4 times 3 times 2 times 2 1 times X to the negative 5 and now you can kind of start to see the pattern

it starts to become really obvious if I'm trying to find the nth derivative of F at X what happens is well first off we know that this f of X is our n equals 0

term this is our n equals 1 term this is N equals 2 this is N equals 3 okay that's how we figure out because we we see if we have an alternating a negative here so we want to know if we

need to put N or n plus 1 in the power ok you see that the odd terms are the negative ones so this is just going to

be negative 1 to the N ok now next let's take care of this factorial that we see here here we see a 2 times 1 here we see a 3 times 2 times 1 here we see a four-ten-four 20 times 2

times 1 okay now you can kind of say that we have a 4 factorial on our third derivative a 3 factorial on our second derivative so it's always one more and

then factorial so for our nth derivative it's going to be n plus 1 factorial okay

so lastly lastly we have our X ok our X to a power now for our zero derivative and what which is just f of X we have negative 2 for a first derivative

negative 3 for a second derivative negative 4 ok so it's always it's always two greater but in the negative direction so how we write this we can just put it on the bottom actually and

get rid of the negative we can write it as X to the n plus 2 okay so now the

second step step number two is to plug in your x equals a ok so plug in a for X and our a is negative 1 as it says in

the problem so F of negative 1 that's going to just equal 1 F prime of

negative 1 will equal to f double prime of negative 1 equal 2 times 3 F triple prime of negative 1 will equal 2 times 3

times 4 all right and you can kind of see pretty much you what's going on here already

ok so if you have a few of your nth derivative of F at negative 1 you will get your n plus 1 factorial still that's

not gonna go away ok you'll have well you'll have on the bottom here ok you'll have a negative

when to the n plus two okay it'll be negative one to the n plus two why will you have that well the reason why you have that is because you're plugging in

and a negative one into your X so what's happening here and of course you have that negative one to the N as well okay what's happening here is that this is

for all odd terms okay and this is for all terms so you've two negatives contributing on all our terms so really this isn't going to matter at all so all

you're getting on your unter imita v' of f at negative one is literally that n plus 1 factorial and that's it so moving on to step three which is you just set

up our series now we know the format for a Maclaurin series are not mcclure we're on taylor for taylor series is the nth

derivative of f a which in this case is negative one over n factorial times x

minus a to the nth power now we can plug in what we have here this n plus 1 factorial in for our F derivative of F

at negative 1 right that's what we found that's why we found it to plug it in here and we put that over n factorial and we multiply it by X minus a to the eighth

power okay well we know that that n plus 1 factorial that's the same thing as n plus 1 times n factorial the n factorials cancel out and you're left

with that n plus 1 times X minus a to the eighth power and I forgot substitute in here we just get now our a is

negative one okay so we get X plus 1 to the N power all right so now that's really it that's our series and sum from

N equals zero to infinity of n plus 1 times X plus 1 to the eighth power all right so that is going to do it for our calculus 2 final review all right

good luck on your exam I really hope this helped if you want some more of you remember I have that hundred question review video down in the description ok it's like 2:30 in the morning right now

and I'm doing this for you guys alright so if you want to consider supporting me on patreon I'm gonna add you that was a little guilt trip right there so you know consider supporting me in the form

of currency on there and yeah so good luck on the final exam I hope you do well and that's going to do it for this video