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[Music] what is an anti-derivative what is an

indefinite integral maybe you’ve heard of them

their reputation does often precede them well can you answer this question can

you find those functions whose derivative is equal to a given function

in this video you’ll also learn how anti-derivatives

lead to finding areas hi everyone i’m dave welcome back to the calculus one

explore discover learn series in this video i’m going to cover anti-derivatives

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first and then i’m going to talk about indefinite integrals and then we’re

going to see some applications of these ideas

and then stick around to the end where i go over some

exercises let’s get started what is an anti-derivative so we’re going to go over

anti-differentiation by defining an anti-derivative function

and working out some examples on anti-derivatives

so we’re also going to concentrate on the following problem

if capital f is an antiderivative so i’m going to use capital f often to denote

an antiderivative of a continuous function

so notice there’s a uh the case matters here

i’m going to use um lowercase f for a given function

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and a capital f for an antiderivative now any other antiderivative of

of a given function f must have the form capital f of x plus c where c

is some constant and we’re going to show that if we have an

antiderivative function then if you add a constant to that

we’re also going to have an anti-derivative

and we’re going to use the mean value theorem here

to show that all anti-derivatives must have this form

that is an antiderivative plus a constant so these are the four main goals

of this video here and we’re going to go through them here

and and plus a lot of examples lots lots of examples so that

all of this here given to you up front right here makes sense that’s the point

of this video is to understand all that now in order

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to really get the full value out of this video

is you need to have uh watch the mean value theorem video

so that one is sort of a linchpin and the idea here that antiderivatives

is a family of functions and and why exactly is that true

so um let’s get started we’re going to start off by first of all

what is an antiderivative so a function is called an

antiderivative of a given function okay so you’re given a function and that

function that you’re given is i’m going to call it lowercase f so

we’re given a function and an interval and then we’re going to say oh

i have an antiderivative of that function so this is similar to

when we’re talking about derivatives right so in order to know what a derivative

is first you’re given a function right and then you can find the derivative of

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that function so we have an anti-derivative

capital f if you take the derivative of the anti-derivative

you get back to the original function f so that’s what that

first sentence is saying right there so you know you have an anti-derivative

if you take the derivative of that anti-derivative and you get back to your

starting place your given function f and so we’re going to say here

um you know let’s try to get comfortable with this definition here before you

before we move on here so let’s say our original function that we’re given here

is an x squared [Music] all right so we’re given a function how

do we find the antiderivative how do we find antiderivative

and i’m going to use the uppercase f what is the antiderivative equal to

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what is it and you know to answer this question we need to

know that you know whatever we say this is we have to be able to take the

derivative of it and get back to the original function

so we’re asking the question what function do we take the derivative of

to get an x square so f of x you might think of something like an x to the third

is that is that an anti-derivative if i take the derivative of this will i

get an x squared let’s try and see what is the derivative of an x to the third

it’s three x squared right so th these are not the same

so this is not an anti-derivative it’s close

but we got the three in the way so what about if we divide by three then

how about this is this an anti-derivative well the question is let’s take the

derivative so if we take the derivative of this we get

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the 3 over 3 and we get an x squared and so yes this is an antiderivative

because if we take the derivative of it we get back to our original function

so here we have found an antiderivative of x squared

what about if we have something like how about um something like

just an x what is going to be the antiderivative or what is going to be

an antiderivative of just an x so we’re asking what is the function we

take the derivative of to get an x so you might guess something like

x squared is this an antiderivative well let’s see if i take the derivative

we get 2x that’s not the back to the original

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there’s a 2 in the way so how about we divide by

2 or or let’s say multiply by a half so is this an antiderivative

and let’s see what’s the derivative now the 2’s cancel

we get back to an x right so this is an anti-derivative this function

right here is an anti-derivative because

if we take the derivative we get back to x

now notice that what about if i add a 5 here

what is this also an anti-derivative the answer is yes because

if we take the derivative of this we still get an x

because the derivative of the five is zero right

so in fact that’s why we always you know we’re always careful and say

here’s a function and here i found one anti-derivative

in fact this is also another antiderivative or we could say minus three here’s

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another antiderivative because this has the same

derivative as the original function so finding an anti-derivative is sort of

like reversing the question um you know and we we study derivatives

for so long in calculus one that we know it

here’s the function find the derivative now

we’re asking reverse here’s a function and now we’re asking find a function

whose derivative is equal to the original function here

all right so hopefully that gets you a little bit

better understanding of what an antiderivative is

and the the word the crucial word here is not only antiderivative but

in anti-derivative so we’re looking for an anti-derivative so because you could

have lots of anti-derivatives here’s the function here’s one of the

anti-derivatives is this an anti-derivative well what’s the derivative

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2x minus 3 do i get back to the same no so this one is not an antiderivative

but if i just left the 3 here now the derivative of the minus three zero

yes this is an antiderivative right here [Music] okay so if

we have this um important theorem here now which says that

if you have an f antiderivative of a function then every antiderivative

must have this form here which is an antiderivative

plus a constant so in other words uh this example that we’re looking at

here a second ago when i have x squared here and i said oh i found an

antiderivative which was um you know one half

x right um what was an antiderivative of of of f here is this one

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well what’s the derivative of this right so the derivative is so that’s not an

antiderivative right so how about one third x to the third

is this an antiderivative of x squared so we take the derivative and we get x

squared so yes this is an antiderivative now

we also said a minute ago that if we add a constant

we still get the same derivative so this is also an antiderivative of x squared

this this function right here is also an antiderivative

well how do we find all the antiderivatives

what if i have a 50 right so in other words we can add a constant to the

to an antiderivative and still have an anti-derivative

but what this theorem is saying is more impactful

it says that yes you can add a constant to an antiderivative

and still get an antiderivative but it’s also saying that

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if you take any antiderivative it has to look like an anti-derivative

plus a constant so let’s see why this is true so we’re going to do this

theorem here in two parts so i’m gonna call this one here part one

part one and what we’re gonna assume for this first part here

is that if we have a function here g of x and it looks like

the anti-derivative plus c [Music] for all x on an interval and if we know that

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capital f is an antiderivative for all x and c rolex and i and c is any constant

see it’s any constant that’s the word constant here so

if we have all these three things here if we have an antiderivative capital f

is an antiderivative then if we add any constant to it

then that will still be an antiderivative is that true

let’s see so the derivative of g to find out if this is an antiderivative of not

or not to find out if g is an antiderivative i’m going to take the

derivative of g so the derivative of g is just the derivative

of f plus c and the antiderivative that we know of

and we add a constant to it is this still an antiderivative

well how’s the derivative work right it’s the derivative it’s f prime

and then plus the derivative of a constant is zero

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we know that this is an anti-derivative so the derivative

of the antiderivative is back to the original function here

so what we just showed is that if you have an antiderivative

and you add a constant to it you still get an antiderivative

capital g is an antiderivative because the derivative of

g is back to the original function here f and this happens for all x and i

hence g is also an antiderivative an anti-derivative and anti-derivative of

f so what this is saying is that if you take any anti-derivative of f

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and you add a constant to it then that and then then that expression there

right there is also an anti-derivative so that’s pretty nice once you found one

anti-derivative you can add a constant to it and you’ll

still have another anti-derivative so that is the intuitive part of this proof

and now we have to have the second part which says

that no matter what antiderivative you have it has to look like another

antiderivative plus a constant so let’s see how to prove that so part two

call this one here part two it’s the same proof but i want to divide it into

separate parts so that you can get the feeling for how the proof should flow

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all right so now we’re going to assume that g is any antiderivative is any

anti derivative of f on i now i’m going to make up this function here capital h

which is just going to be g minus f [Music] and that’s you know for all x on the

interval so in other words we know of one anti-derivative

and this function right here capital g is

we’re assuming is is any antiderivative is any other antiderivative

and so we’re looking at this function here h now for all a and b

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in the interval i for all a and b and i h is continuous

on the interval i uh interval a b so we’re randomly choosing we’re just

arbitrarily choosing we’re not putting any conditions on the

a and b other than we’re choosing it in i and once we do that we know that h

is continuous on a b closed a b and it’s differentiable [Music]

is differentiable on a b so we’re going to use the mean value theorem

so by the mean value theorem we know that the derivative of h

is equal to the difference of h on the boundary over the difference of a and b

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for some c and a and b for some c in the interval a b

this is exactly what the mean value theorem gives us

because we know it’s continuous on the closed and differentiable on the open

so we can use the mean value theorem here to get some c

in this interval here where where the derivative of

h is equal to this expression right here okay good so now let’s notice

notice that the derivative at c or in fact the you know the derivative at c is

is what is the derivative of g at c minus the derivative of f at c

um but these are equal to f of c minus f c which is all in fact just zero

because remember these are anti-derivatives so the derivative of

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g is f and the derivative of f right because these are

anti-derivatives so their derivatives is equal to the original function so if

i take the derivative of capital g and i plug in a c

that’s the same thing as plugging in c into f and the same thing here f is

f is our known anti-derivative and g is just an arbitrary anti-derivative

but any case they’re both anti-derivatives so they’re both equal to f of c

and so we’re going to subtract we get 0 here so what this tells us is that

in fact this right here has to be 0 right here this is 0 and

so these are this is not zero so this has to be zero right here so

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h of a equals h of b so that says h of b which means h is constant

h is constant on the interval here a b or in fact i because

remember we chose a and b here for all a and b and i no matter what we

chose in i a and b that these values are the same

so h has to be constant on the whole interval so g of x is equal to

or let’s say here uh right so h is constant right so

so what is h h is just simply g minus f right so g minus f

is equal to some constant right and so thus g of x is equal to f

of x plus c where c is some constant and that’s exactly what we needed there

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[Music] so if we take an arbitrary anti-derivative

it’s going to be equal to our known antiderivative plus a constant

so any antiderivative must have this form a known antiderivative plus a constant

all right so there’s that theorem there there’s the proof of it it relies upon

the mean value theorem here and that’s sweet okay so let’s go on part two

[Music] okay understanding the anti-derivative uh now much better

now we’re ready to talk about indefinite inter integral

so when we look at this expression right here

the right hand side we just talked about capital f of x

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plus c where c is a constant now the left-hand side is the new

notation that we that we’re using right here

so this notation right here the integral of f of x

is the anti-derivative here so this is the anti-derivative

and this is indefinite integral this right here is the integrand

so here we’re given a function this is given to us

and we want to find an indefinite integral and it’s equal to the antiderivative

plus c [Music] and this right here is the the

differential right here that’s important this says what we’re taking the uh

anti-derivative with respect to with respect to x and so it’s important to

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have this terminology downright and this is a family of functions right here

right uh capital f of x is a is an antiderivative

and then we add a constant to it and that’s called the indefinite integral of

the given function f of x with respect to x and it satisfies the condition right

that the derivative of the antiderivative

is back to the original function for all x in the domain so it’s

important to remember that this represents a family of functions because c

is an arbitrary constant c could be zero c could be two c could be negative

square root of two c is an arbitrary constant so the process of finding all

anti-derivatives is called anti-differentiation or

indefinite integration so there’s two types of

integration that we’re going to talk about in calculus one

indefinite integration which we’re talking about first

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and then we’ll talk about definite integration later

so in this video we’re talking about indefinite integration

and then in the next episode we’re going to talk about

uh substitution with indefinite integration

and then we’ll put it all together when we talk about definite integration

okay so here’s a first example here find um the family of anti-derivatives

of this function right here so the way that we find the family of

anti-derivatives is we find one of the anti-derivatives and then we find

and then we add the constants to it because that theorem says that all

anti-derivatives must look like one of the anti-derivatives plus an arbitrary

constant so we just need to know one of the antiderivatives

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so what’s an antiderivative of cosine so an antiderivative of cosine you might

remember is right so what’s the derivative what do you take the derivative of

to find sine right so f of x will be equal to minus cosine x

and let’s check that what is the derivative of the answer

of an antiderivative right so we have a minus sign and the derivative of cosine

is minus sine so we’re going to have a negative times a negative

which is equal to sine x so this says right here

so this is an n is an antiderivative is an antiderivative of f

minus cosine is an antiderivative so how do we know my

minus cosine is an antiderivative because if i take the derivative

of it if i take the derivative of minus cosine i’m going to get minus and the

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derivative of cosine is minus sine and that all comes out to be sine which

is our f of x so the family of anti-derivatives the family of anti-derivatives

of f is the minus cosine plus c now how do we write that right minus cosine

plus c we can write that using indefinite integral notation so the indef

the the indefinite integral of the the sine function the one we’re giving dx

is equal to the family of anti-derivatives minus cosine

plus c so this is the notation right here so this is saying the indefinite

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integral of sine x is minus cosine x plus c what that means is if we take the

derivative of this right here we get back to sine

all right very good let’s go on and try our next one

how about x to the minus three so let’s try again

so first off we need to find an antiderivative not going to be minus cosine

and let’s try and see what happens so what is the

function that we need to take the derivative of to get back

to x to the minus 3. so we need to try to reverse

taking the derivative so normally when we take the derivative

we’re going to do what right what do we do when we take the

derivative of x to the minus three we we bring the power down

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and then we subtract one so minus four so how do we reverse this process

right this is derivative anti-derivative is reversing the process

so instead of bringing the power down and then subtracting one

first i’m going to add one to the power and then divide instead of multiply i’m

going to divide so we’re going to do that process there

so if i add 1 to x to the minus 3 it’s going to be minus 3 plus 1 so it’s

going to be x to the minus 2 and then i’m going to divide by minus 2.

why is that why does that work let’s see if that works what is the derivative

what is the derivative here so minus 2 comes down and cancels

and then i do x to the and i subtract 1.

so the derivative x x to the minus three and that’s exactly what we needed that

is f of x so this is an anti-derivative right here because if i

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take the derivative of it i get back to my original function here

so here’s an antiderivative so what is the family of all antiderivatives

is just find one of them and then add constants

now what’s a better way of writing x to the minus two over minus two

let’s just say it’s minus one half and then so let’s say it’s minus one over

two x squared and then plus c to find all of them

we just simply add the constant to them so let’s write that with indefinite

integral notation so the indefinite integral of 1 over x to the third

right that’s x to the minus 3 is equal to don’t forget your differential

is equal to minus 1 over 2x squared plus c so here’s an equation right here

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in indefinite integral notation the indefinite integral of x to the minus 3

is equal to minus 1 over 2x squared plus c now how do we know this is right

if we take the derivative of this we get 1 over x to the minus three and you can

check that because because you know how to take derivatives

if you take the derivative of this you get back to your starting place here

okay so there’s our second example there um how about if we just do an arbitrary

n instead of looking at say minus three what about if we look at an

n for example what if your function is x to the fifth or your function is x to

the three halves any n you want except minus one

so how do we find an antiderivative of this

so i’m going to do the reverse of the derivative process

i’m going to find an anti-derivative so instead of

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bringing the n down and multiplying and then subtracting one

i’m going to reverse all those so i’m first going to

add one and then i’m going to divide by it

and here’s why we can’t use n minus one here because that will make that zero

there down there so this is an antiderivative how do we

know it’s an antiderivative let’s find the derivative of the anti-derivative

so if i bring the n plus one down they’ll cancel and then i say in

x to the m plus one minus one which is just x to the n and that’s back

to our original function so this is an anti-derivative so is an anti-derivative

again how do you know something is an anti-derivative you take the derivative

of it and you get back to the original function

so this is an anti-derivative so now we can find the family of anti-derivatives

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so the integral the indefinite integral of x to the n is equal to

an anti-derivative plus a constant so this is the family of anti-derivatives

is one way of saying it or this is the indefinite integral of x to the n

is equal to the family of anti-derivatives where c is an arbitrary constant

arbitrary constant or you could just say where c is any

constant or you could just say where c is a constant they all mean the same

thing it sounds fancier to say where c is an arbitrary constant

but they all mean the same thing where c is any constant

or where c is a constant okay so there’s the in indefinite integral notation

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and there’s our family of anti-derivatives and again i know that’s right because

if i take the derivative of this i get back right here with x to the n

okay so let’s look at next one um now now that we’re kind of getting a

better feel for this process because we’ve taken so many derivatives

up to now in calculus one that you can go back and look at all of

the derivatives that you found and you can make new antiderivative

formulas for them so for example um what are some of the um derivative

theorems that we know of for example we know about the derivative

of the constant you can pull out the derivative of a constant

so you can do the same thing for anti-derivatives

we know how to do the sum rule how do you take the derivative of a sum

you take the sum of the derivatives and so we can do the same thing for

anti-derivatives the difference say same thing

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and putting them all together we have the linear rule

if you take the anti-derivative or the indefinite integral of a linear

combination it’s going to be the linear combination of the antiderivatives

okay and now we have this um this power rule that we’ve been working on

and we have some very familiar um and indefinite integral formulas here

so what is the you know how do i know this one right here is true

that the indefinite integral of e to the x

is just e to the x plus c if i take the derivative of this

i get back to e to the x right what’s the derivative of e to the x

plus a constant well the derivative of the constant is zero

and the derivative of e to the x is e to the x

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so the indefinite integral of e to the x is e to the x

plus c what about next one well we did this one right here in a

previous example right the indefinite integral of sine is a minus cosine u

plus c and you can check that what’s the derivative of minus cosine u plus c

about this one the antiderivative of cosine

we can check that that one is true check for

right so i’m going to look at sine u plus c and i’m going to take the derivative

what’s the derivative of sine it’s cosine

what’s the derivative of a constant it’s zero so plus zero so this is a check

this is a check that number four is correct i took the derivative of the

right hand side and i got the integrand back

what about number five so number five we know that the derivative

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of tangent derivative of tangent plus c the derivative of tangent

is secant squared plus derivative of the constant derivative of secant is

secant tangent so number six works the derivative of the right hand side is

equal to cosecant cotangent the derivative of

minus cotangent is cosecant squared ah remember the inverse

formulas we had an episode on derivative of inverses

what is the derivative of sine inverse right so we can check that formula right

there and because we know what the derivative of

tangent inverse is and because we know the derivative of

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secant inverse so those formulas come because we know what the derivative of the

inverse functions are okay so let’s look at some more examples

so we have some basic formulas that we know of

but integration is harder than just that so here’s an example

if we want to look at say this one right here the integral of

number ten right the integral of one plus x squared dx

by looking at number 10 here is just you know tangent x plus c

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or tangent inverse x plus c but this example right here wants us to

look at the integral of the indefinite integral of

x squared over one plus x squared right and so we know what this one is

but we don’t know what this one is and when there’s an x squared here

that makes the problem very very different

what you cannot do is take the x squared out and have the integral of 1 plus

x squared and now we know what all this is it’s just that right so this does not

work that’s just a bunch of junk so how do we do this

we need to find the we need to find an antiderivative of this what do you take

the derivative of to get an x squared plus x squared plus one

and so that can be a little tricky so what we’re going to use is x square

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plus one i’m sorry x squared over x square plus one i’m going to add a one

and subtract a one in order to make this work so

i’ll take the x square plus one right here with the first two

and then i have a minus and the reason why i do that is because

this is a one right here this is one minus so we just we’re trying to work with

this function right here x squared over x squared plus one and i

wrote it down like this and the reason why i wrote it down like this

is because i know how to work with these two right here so here we go

the indefinite integral of x squared over x square plus one

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is equal to so instead of using this i’ll use this one right here so one minus x

squared over x squared plus one okay so this one looks harder to work

with this one looks a little bit easier to work with

and the reason why is because we have our difference rule so i can just work on

this one right here and then minus and then i can work on this one right here

sorry this is a one here that should be a one here right all of this is equal to

the integral of all of that just integrating both sides

so this right here will be the integral of one over x squared plus one

so integrating this finding the indefinite integral of this one

i broke it down into two simpler ones this one is straightforward it’s just x

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and then it’s actually plus a constant now how do we know this is right well

what’s the derivative of this derivative of x is one derivative of constant zero

right so this is the antiderivative right here

minus and then this one right here remember this one was

tangent inverse and then we have a constant

now we have to be a little bit careful about the constants

because this one is an arbitrary constant coming from this

family of anti-derivatives so we could put a c1 here

and this is a constant coming from this family of anti-derivatives right here

so we could put a two here in fact we have a minus here so that

should be minus for the whole thing or if you want uh parentheses

right because we have this one this family of antiderivatives

minus this family of anti-derivatives okay

so you could write it like this where c1 and c2 are

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arbitrary constants or you could just write it like this where c

is an arbitrary constant where c is a constant so c 1 minus c 2

where c 1 and c2 are arbitrary is the same thing as writing an arbitrary

constant over here so this is really unnecessary right here

we could go straight from here to x and then

minus sign and then the and then this right here

is you know the derivative of tangent inverse is

1 over 1 plus x squared and then plus a constant

so i just basically saved up all my constants from both

and put him together as one arbitrary constant

okay so we found this right here and we did a little trickery

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to it so integrating is much harder than finding derivatives

all right let’s look at the next example find the general anti-derivative or in

other words find the indefinite integral and

if i wanted to say find the derivative of this function

we could use a product rule or we could expand it all out and then take

the derivative of each term finding the antiderivative though there

is no product rule we’re going to find the anti-derivative

and the way to do that is to expand it out so we’re going to have

f of x we’re going to rewrite it so i’m going to say one times each of those two

so that’ll be 1 minus 4 over x squared and i’m going to say 1 over x times each

of those two so 1 over x and then minus 4 over x to the third

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so we can write f of x like this and in fact we may want to write it down as

1 and then plus 1 over x and then minus 4 over x squared

and then minus four over x to the third maybe you’d like to write it like that

in fact maybe you would like to write f of x like this

one plus x to the minus one minus four times x to the minus 2 and

then minus 4 times x to the minus 3. in any case however you want to write f

we have to go find the antiderivative of it

all of them so the antiderivative or the indefinite integral of f will be

all of this one plus x to the minus one minus four x to the minus two minus four

x to the minus three okay good now i can go and look at them

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separately one individually one by one and we could write all that out

in fact you should see it at least once so i’m going to write it out and then

i’m going to erase it because i don’t think that you should write it out

but i think that you should visualize it so how can you visualize it if no one

has ever done it for you at least once so this is one dx

plus x to the minus one dx plus minus four x to the minus two dx and then plus

and then the last one here minus four x to the minus three dx

and then maybe put parentheses here all right so all of that

so we’re gonna go and enter and find the indefinite integral of

each one of these four here so what is this one right here

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it’s just x plus c plus some constant i’m going to save up all my constants to

the end so x plus c1 okay so i’m just gonna find the

indefinite integral of the one and then this one and then this one and this one

so here we have now this is a minus one here so this is going to be

the natural log of the absolute value of x plus a c2

and then this one right here is going to be i’m going to add 1

to the power and then divide so i’m going to get minus 1 so i’m going to say

minus 4 x to the minus 1 and then divide plus another constant and then plus

and then this last one right here is going to be minus 4 and then

x to the minus 3 plus 1 so minus 2 over minus 2 and then plus a constant

00:46

so what are we getting for all this we’re getting a natural log of the

absolute value of x plus plus x and then what’s this right here plus

4 over x and then this one right here all becomes um 2 over x to the third

or squared and then plus c [Music] okay so this is our general antiderivative

or this is the indefinite integral of this function here f

and you could you could you could test that by taking the derivative of all

this right here taking the derivative of all this right here and you’ll get this

function right here f now to be honest you don’t really need to do each of

these steps sometimes you do sometimes you don’t

just depending upon what they’re asking you but if this is your 200th

00:47

example then you can move here much faster

the first time you see it you could see all these steps here c1 c2 c3 c4 they’re

all constants and then you just put them all together

into one constant and then you just say where c is a constant c is a constant

okay so can you go from right here to right here how many examples

how many exercises will you need so that you compute the one it’s just the x

the x to the minus one is just the natural log of the absolute value

and then the you know use the power rule there

or indefinite integrals to get each of those okay anyways next example

now in this example here we’re going to this is called an initial value boundary

00:48

problem and it’s basically we’re going to reverse the process of differentiation

and we’re going to find the constants so to work on this one right here

we’re given this second derivative and we need to find the function f

so you’re used to doing the opposite here’s the function f

find the derivative find the second derivative

now we’re given the second derivative and we’re asked to find the function

so we’re going to find f of x f prime by integrating so this will be

x plus the square root of x dx and what is this so

when we integrate the x we’re going to get x squared over 2

and when we integrate the square root of x right so this will be x to the

one-half power but i’m just looking at this part right here

00:49

the square root of x so how do we do that so we’re gonna

add one to the power so one half plus one that’s three halves and then i’m going

to divide by three halves plus c right in other words this will be

two x to the three halves over three two thirds so let’s write that up here so

when i integrate the first one the x is just x squared over two

right the power once we add one to the power and didn’t divide

we do the same thing once one half so it’ll be x to the three halves

over three halves and then it’ll be easier just to write it like that

two thirds so this will be two thirds x to the three halves and then plus a

constant all right so so far so good so we know what the second derivative is

they give that to us and now we know what the first derivative

00:50

is well we don’t know what the first derivative is entirely

because c is an arbitrary constant so what they do is they give us a boundary

condition this is called the boundary condition

we know that f prime at one is two so if i plug in

1 into the derivative i should get out of 2.

so let’s do that if i plug in a 1 into the derivative

i should get out of 2. well what happens when we plug in a 1 into the derivative

this is the derivative so what happens when we plug in a 1 so

we’re going to get one half plus this will be two thirds

plus a c so solving this right here for c so we have to move the one half over

move the two thirds over and we get so c is 5 6 when you solve that

right just do two minus a half minus two thirds

00:51

you know just put all that over six multiply by

you know six top and bottom you get twelve

and then minus three and then minus two so right you get five six

in any case our first derivative we have all the way pinned down now

the first derivative is all of this plus the c we just found

using this condition right here the derivative at one is two

we’re able to use the derivative at one is two by plugging in one

we found the c so the first derivative is uh let’s say it say it’s one half x

squared plus two thirds x to the three halves power

plus our constant which is five sixths so it asks us to find the original

function instead we found the derivative well we can integrate one more time

00:52

so to find f of x i’m going to integrate the first derivative

so two-thirds x to the three-halves plus five-sixths so i’m going to

integrate all of that to find the original function here

so here we go one-half x-squared so what do we get

so i’m going to increase the power by 1 and divide so x to the third

over 3 but i already have a one-half here so this is going to be 1

6 x to the third less now this is going to be right so we’re

going to add one so to be it’ll be uh you know three halves plus one

so that’ll be five halves over five halves but we already have a

two thirds in the front of here right so so what is all that when i try

to when i try to integrate this piece i’m using the power rule

so three halves plus one and then divide right so this will be two thirds

00:53

times so if i divide by five half that’s the same thing as multiplying by two

fifths and then x to the five halves so this is what four fifteenths

x to the five halves so sometimes you might need to do some scratch work over

here but if you’re doing the integral of two thirds

x to the three halves all of that right there

is this right here so i’ll put it over here four fifteenths x to the five halves

and then now the last term so we integrated the first one we got that

we integrated the second one we got that now we’re going to integrate the 5

6 and that’ll be 5 6 plus a constant so 5 6 plus a constant 5 6 x

right the derivative of 5 6 x is at 5 six plus a constant now i’m gonna have

00:54

another constant here and i want to confuse it with my previous constant

my previous constant was c so i’m gonna use a constant k capital k there

so in order to find out the function the final answer

i need to find out this capital k here now they tell us that if i plug in one i

get out one so let’s do that if we plug in one here

we get out one now if i plug in a one here i plug in a one everywhere here and i

should get out one so let’s plug in one plug in one here i get one

six plug in one here i get four fifteenths i plug in one here i get five six

and then if you add all that up you get right you get 1 minus 1 6

00:55

minus 4 15 minus 5 over 6 so when you find that k it’ll be minus 4 over 15.

so we have our final function now all of this plus our k which we just

found so it would be 1 6 x to the third plus 4 15 x to the five halves

plus five six x plus a constant which we found

by this initial condition here f of one equals one

and that was minus four over fifteen so there’s the function there

we know this works we can check it if you find the second derivative of this

function right here it will be x plus square root of x and

if you plug in one into this function you should get out one

and if you plug in one into the derivative of this function

00:56

you’ll get out two so this is our function right here so

that’s pretty sweet there’s the work there we start by

integrating to find the first derivative and then we integrate to find the

function we have to find the missing information here the constant but

in the end we get the final function there

okay so let’s move on to the next part applications of the indefinite integral

okay so our first application is going to be finding area

so this is an application to mathematics itself of the process of antiderivative

so let’s read this theorem here and try to make sense of it

it says that if we have a continuous function

00:57

and that function is positive or zero on a closed interval a b then the area

bounded by the curve the x-axis and the vertical lines

so all that bounces certain region and to find that area of the region

you use the antiderivative of f and so let’s look at a

just a sketch of what that would look like so we have here um

a continuous function it’s positive it may touch

down to zero but let’s just draw one here like that f that’s the graph of f

and we got an a and a b and i’ll just draw it in the first quadrant here

00:58

just to illustrate it so the area bounded by the curve which we have right here

the x-axis right here and the vertical lines x equals a and x equals b

we’re bounded by those vertical lines there so here’s the region that we’re

bounding right there and now we have x equals t so let’s move this

to a t here instead of a to b we’re looking at well let’s put a b

here because we want to view this as a function of t

so i’ll erase some of this and say here’s the x equals t

vertical line and so let’s choose a t somewhere in here

and so here’s the area here and we’re going to denote the area function by

00:59

a of t [Music] is the area of the bounded region

we know that it’s bounded because f is continuous and we have these two

vertical lines and we know that the function f of x stays above the x-axis

and so we’re also bounding it below by the x-axis

so here’s an arbitrary t between a and b so as you move t back and forth you’re

going to get different area and so this is a function of t here

when t is actually equal to a well then your area is zero

because you don’t have any width to your region here

and if t comes out to be equal to b then you have the whole area of the

region so we’re thinking about t being the variable and a of t is the

area the bounded region and what we’re saying is that this theorem is claiming

01:00

is that the area is an anti-derivative so this is an anti-derivative of f

so here’s the function f and this area here is the anti-derivative of f

so well when i say anti-derivative i mean with respect to t

on a b so we’re given f of x and so we just use t as the

variable and then we have the anti-derivative

so let me show you an example a concrete example here

so find the area under the parabola over the interval

0 1 and to do that we’re going to use this

theorem here finding an anti-derivative so here we have y equals x squared

01:01

and we’re on zero one [Music] and so we can go back here and say um

look at this theorem here with this example right here we have x squared and

we’re on zero one so what’s going on here so we have

y equals x squared looks like that we touch zero right there

but we’re continuous and let’s say one is about right there

so we have this right here and then we have zero one

so we have x equals zero vertical line and we have the x-axis

so we’re looking at this area in here and let’s pick a t between zero and one

let’s call this right here t so how do we find this area here a of t

01:02

is an antiderivative so what is an antiderivative of x squared

right so we know after having done some practice

that the indefinite integral of x square dx

is equal to x to the third over three plus a constant right so we have

here a of t is we can do this with t now t squared dt

and this will be one third t to the third plus c

and how can we find c well we know that a of zero is zero in other words when t

is zero we’re back here we have no width then the area is zero

so when we plug in zero when we plug in zero here we should get out zero

01:03

so a of zero is one third zero squared plus c so

c is just zero so we found a of t a of t is c zero here one third t to the third

so that is the area so for example if t is one half

then we will have found the area under the curve

you know but only going to one half the midpoint now the problem asks us um

what is the area on zero one so that means t is just one so a of one

is just one third one to the third or in other words just one third

so one third unit squared or you know so that’s the area of the graph

of x squared on zero one if you went all the way

and t was equal to one you would just plug in 1 and you would get the area

01:04

so by finding the general anti-derivative here

with the c but being able to find the c so now i have the area function right

here and you know we can use any value of t between zero and one for example we

can find out one half we would just plug in t’s one half or whatever [Music]

okay so that illustrates that theorem there um next application

so let’s see here a ball is thrown upward with a speed uh 48 feet per second

and from the edge of a cliff 432 feet tall and so let’s see here find its height

above the ground two seconds later when does it reach its maximum height

01:05

when does it hit the ground so let’s try to answer all those questions there

so the motion is vertical and we choose the positive direction to be upward

at time t the distance above the ground is s of t

so let’s say s of t this distance um above the ground [Music]

and let’s use v of t is the velocity and a of t to be the acceleration

and um we’re going to use a of t which is the derivative of the velocity

01:06

is minus 32 because this is taking place on earth so two times

okay so how do we find the velocity then all right so we’re going to take the

derivat uh the integral of the derivative so the velocity will be [Music]

taking the indefinite integral of this right here and so this will be minus 32

dt and this will be minus 32t plus a constant

we can always check that taking the derivative we get back to the minus 32.

now can we determine c what is the velocity at they give us that the

01:07

initial speed is 48 so v of zero that’s a zero is 48 so

you know when we plug in a 0 here minus 32 times 0 plus a constant so [Music]

c is 48. in other words our velocity velocity function is minus 32t

plus our constant 48 so there’s our velocity

it’s you know find the height above the ground

when does it reach its maximum height so on so we can see that

the ball reaches the maximum height when the velocity is zero

when the velocity becomes zero that’s when it reaches its maximum height

01:08

so we’re going to look at that now and say v of t equals 0 which is

minus 32 minus 32 t plus a constant minus 32 t

plus the constant so we need this to be equal to zero

so when we do that t is going to be equal to move the 48 over divide by minus 32

and we’re going to get t is you know 48 over 32

or 1.5 and we’re going to go with seconds here

so that’s when it reaches its maximum height

it’s going at 48 feet per second so this is seconds

so that’s certainly the time there when it reaches its maximum height

so we’ll just say here max height at t equals 1.5 seconds

01:09

okay so now we want to find the s of t to answer some more questions

so the s of t is going to be the think of it as position right so it’s

going to be the um so v is the velocity there so

s of t is going to be the integral of the derivative here minus 32 t plus 48.

so we’re going to be able to find the position function

and we’re going to get here so add 1 and divide so we’re going to get

minus 16 t squared plus 48 t plus a constant now i already use c as a

constant down here so i don’t want to get it confused i’ll say k perhaps

01:10

and we’re giving some initial conditions here so s of um so

we’re given here 432 right so f of 0 is 432 so when i plug in 0 i plug i get a 0

here and actually that should be 48 t right plus a constant right

so again check derivative of this we get back to the original

so when i substitute in 0 here i get 0 0 and k so when i plug in 0

i get 0 plus 0 plus k so k is just 432 okay good so now we have our s of t

01:11

now it’s going to be minus 16 t square plus 48 t plus our constant

so given any point in time this is our position right here so

the height of the function is given by this right here right that’s the position

so what is the ball hit the ground so s of t

equals zero when the height is zero so we need to solve all of this

equal to zero so when we solve this equal to 0 right here

well we can use the quadratic formula here and we’re going to get t is equal to

3 plus 3 square roots of 13 over 2 which is approximately 6.9

seconds so 6.9 seconds later it’s going to hit the ground so we found the height

01:12

function right here to answer the question find its height above the

ground two seconds later right there and when does it hit the ground

when the height is zero and then so yeah using the quadratic formula there

and then we get uh about 6.9 seconds later

all right very good so there’s you know some elementary position velocity ball

type of problem now let’s look at something like some you know economics maybe

okay so now we have a manufacturer estimating and

they want us to find the domain demand function

so we’re given the marginal revenue given the marginal revenue so let’s find

the revenue first knowing the marginal revenue we’re going to find the revenue

01:13

so the uh so the revenue is going to be the integral of the marginal revenue

and so this is going to be the integral of 240 plus 0.1 x

dx so in order to find the revenue knowing the marginal revenue we’re going

to find the indefinite integral and we have 240x

plus so we’re going to increase and then divide so divide that by 2

so we’re going to have 0.05 x squared and then plus a constant so that gives

us the possible revenue um can we find the constant [Music]

01:14

so you know the revenue is zero when we have produced zero units right so

well if if we substitute in a zero we should get out zero here

so let’s substitute in zero so put a zero here and a zero here

we’re going to get zero plus zero plus c you know so

this is zero the revenue r of zero is equal to zero

so c is zero okay so the revenue function is just 240 x plus 0.05 x squared

you know plus zero so there’s the revenue function there

now the domain the demand the the sorry the demand function

is the revenue divided by the number of units

or think of it you know differently the revenue

is the price or the demand function times the number of units right so to

find the demand function you’re just going to divide by the x

01:15

all right and so we just found the revenue so we just need to divide it by

x so we’re going to get 240 plus 0.05 x so that’s the demand function there

all right very good um and so if you’re not like um clear on the domain function

and the revenue and all that stuff in our last video

i’m sorry last video was the hopital’s rule the previous

uh video on the applied optimization we went into that in more detail

this is kind of a follow-up on all that um

of course in that first initial video we just knew derivatives

now we’re applying anti-derivatives so given some information uh before

we were given the demand function and we’re asked to find something like the

01:16

marginal revenue which with it so this is kind of the reverse of the problems

we did before okay so let’s go to the next one here [Music]

okay so an excellent film with the very small

advertising budget must depend largely on word of mouth advertising so in this

case the rate of which weekly attendance might grow

is modeled by this function right here this is a derivative where t is

in the in the time in weeks since its release

and a is in the attendance in millions so we want to find the function that

describes weekly attendance in other words we want to find the function a

01:17

and then we want to use that function right there to make some

predictions so how can we find the a knowing the derivative of a

so we’re going to say that a of t is the indefinite integral of

d of a of t right dt so we need to find out what this is

and this is our a function our our d a right here is actually minus 100 over

c plus 10 and we have a square on all that and then plus 2000

all over t plus 10 to the third to the third and then so we need to find

01:18

the antiderivative of all this i’ll just write like that so the d a

right we have right here i just use the d a so now i want to find the

antiderivative of each of these here [Music]

okay so when i’m looking at this first part here um

i’m thinking about this like this i’m going to write it down here so it’s 100

sorry 100 and then what is this right here it’s t plus 10 to the minus 2 right

so i’m going to add 1 and divide so it’s minus 1 and i’m going to divide by it

so i’m going to get 100 and then t plus 10 right because it when i add

what’s gonna be minus one so it’s going to be in the denominator here

and then minus and then i’m thinking about this one right here is

2000 and then t plus 10 to the minus 3 right and so i’m going to add 1 and then

01:19

divide so i’m going to be chopping that 2 000

in half so i’m going to get minus 1 000 and then t plus 10

squared and then i get some arbitrary constant let’s call it c so

this is my a of t here is all of this now if you’re not sure about that take

the derivative of this you take the derivative of this right

here you get this right here and if you take the derivative of this

right here t plus 10 right there you take the derivative you get this

right here and then the derivative of the constant is zero

okay so we found our a of t he’s in time in weeks since its release and a is

in millions of of people so now when t equals zero then we can find the constant

01:20

so when the what’s the attendance when there’s no time that has passed right

so this will be 100 over 0 0 plus 10 minus 1000 over 10 squared plus

c and this is just c and so c is zero as it should be c is just zero

now of course when time is not passed your attendances should be zero right

so this is our attendance function here let’s just go and write it out real

quick it’s just 100 t plus 10 minus 1000 here over t plus 10 squared [Music]

01:21

and so number part b here is what’s what’s the

attendance when it’s in its 10th week so we’ll just plug in 10. [Music]

so we get 100 over 20 minus 1000 over 20 squared

and if you work that out you get 5.2 or two point five million people

all right so just plugins in get our number 2.5 2.5 million people there okay so

last example here [Music] here we go now we’re given the marginal

01:22

cost for a product and so sometimes people denote the

marginal cost with that bar over the mc and its fixed cost is 340

and we’re given the marginal revenue is eight adx and we’re looking for find

the profit or the loss from the production and sell of some

different units okay so let’s first find the cost function

knowing the marginal cost we should be able to find the cost function

now also knowing the marginal revenue we should be able to find the revenue

once you know the cost and the revenue you should be able to determine the

profit so let’s get started so i’m going to look at

c of x first the cost and it’s going to be the

indefinite integral of the marginal cost so 60 and then square root of the

01:23

x plus 1 dx so what will our cost be so i’m looking at the x plus 1 with a

square square root so let’s look at it like this x plus one to the one half

right so we’re gonna add one and divide so when i add one i get three halves

and then i’m gonna divide by three halves now we have a sixty of course

that’s constant so sixty and then x plus one to the three halves

and i’m gonna divide by three halves and of course we have a constant i’ll

say constant k um now when we substitute in zero we should have the fixed cost

340 there so cost at zero is 340 and what happens when we plug in

zero what do we get for all this right here

we plug in zero here that’ll be one to the three halves so that’s just one

01:24

so what’s 60 divided by three halves or in other words multiply by two thirds

plus the constant k so we need to find the k now multiply 60 times two thirds

and then subtract that from the 340 and k is 300. okay good so

we have the cost function here now it’s going to be the um

you know 40 one plus x to the three halves plus the constant we just found 300

[Music] um yeah so now we can find the marginal revenue

so the revenue function oh sorry now we can find the revenue

01:25

we know the marginal revenue so let’s integrate and so 80x

and so this will be x squared and then divide by two

so we’re going to get a 40. so 40x squared plus some constant let’s call it c

um now the revenue when we’ve made zero units should be zero

so if we plug in zero here the whole thing should be zero

so c is zero also or in this case just like the last example because r of

zero should be zero when we we have no units we should have no revenue

so if i plug in 0 here 0 here the whole thing should be 0 so c has to be 0 also

so our revenue function is just simply 40 x squared all right so now we’re ready

01:26

for our profit so remember that the profit is revenue minus cost

so we found these two let’s put it together so we have 40x squared

that’s the revenue minus the cost and here’s the cost right here so then we have

40 and then one plus x to the three halves and then this is minus

all of the cost so this should be minus all of the cost so minus this minus that

so minus 300 there so there’s the profit function right there

and we want to find the profit or the loss from three units so

we need to plug in p of three we need to find p of three so let’s just

plug in three everywhere times nine minus forty times what is that four

01:27

so four to the three three halves and then subtract 300 from there and so

for all that you get minus 260 so that’s certainly a loss let’s say dollars

and p of the other one they ask us for here part b

would be p of eight so now we’re going to plug in eight

so forty times the eight squared and then minus forty times the

um nine to the three halves now and then minus the three hundred

and so for all that we’re going to get eleven eighty dollars

okay so there we we we worked it out twice we found marginal cost they gave us

marginal costs we found the cost it gave us marginal revenue so we found

the revenue we put together the profit function and

01:28

then we plugged in some values to find out how much money we made

or how much money we lost all right so very good so let’s look at some exercises

okay so these exercises are for you to try and we have some um

problems here to help you get started down the path of being able to find

anti-derivatives on your own and so here i give us three problems at a

time so number one has an f a g and an h and we want to go find

the anti-derivative of it we want to write out an indefinite integral

equation for each one of these and then we have some more here four

more problems here so i recommend trying those problems out

01:29

on your own thinking hard about them and finding your solution to them

and in the comments below tell me how you did tell me if you did a great job

tell me if you need some more help let me know

and we can make a video over these exercises if you want um so

i want to say thank you for watching and see you on the next video which

is about in definite indefinite integrals but we’re going to do

substitution with them so we’re going to gain a lot of strength

in terms of the number of types of integrals that we can use

and find and so we gotta we got up really close to using the

integration by substitution in this video but

we didn’t do it and we’ll do it next time and i look forward to seeing you then

01:30

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