PROFESSOR: Hi. Welcome back to recitation. We’ve been talking in lecture

about antiderivatives. So I have here a

problem for you. Just an exercise about

computing an antiderivative. So the question is to

compute an antiderivative of this big fraction. So on top it’s got x to

the eighth plus 2x cubed minus x to the 2/3 minus 3,

that whole thing over x squared. So just a quick

linguistic note about why I said an antiderivative

instead of the antiderivative, and then I’ll let you

work on it a little. So an antiderivative. There are many functions whose

derivative is this function. Right? So they all differ from

each other by constants. So I would be happy with any one

as an answer to this question. That’s why I chose the

word an antiderivative. So I’m looking for a

function whose derivative is equal to this function. So why don’t you take a

couple minutes, work this out, come back and you can check

your answer against my work. OK, welcome back. So we were just talking about

this antiderivative here. So one thing you’ll

notice about this function is that I’ve written this

in a sort of a silly form. And it’s probably

a lot easier to get a feel for what this

function is if you break this fraction apart

into its several pieces. So for example, x to the

eighth over x squared is just x– So,

well OK, so let me, this antiderivative

that I’m interested in, antiderivative of x to the

eighth plus 2x cubed minus x to the 2/3 minus 3,

over x squared dx. So I’ve written

this in a silly form and you can get

it in a nicer form if you just realize

that, you know, this is just a

sum of powers of x that I’ve put over this

silly common denominator. So our life will be a little

simpler if we write this out by splitting it up into

the separate fractions. So if I do that, this is just

equal to the antiderivative of well, x to the

eighth over x squared. That’s x to the sixth. And 2 x cubed over

x squared is just x. So I have x– sorry,

it’s 2x– plus 2x. Now, OK so x to the

2/3 over x squared. So that’s x to the 2/3 minus 2. Which is x to the minus 4/3. And minus 3 over

x squared, so OK, so we could write that as minus

3 over x squared, or maybe it’s a little more convenient

to write it as minus 3 x to the minus 2, dx. So far I haven’t really

done anything, you know. A little bit of algebra here. OK, but now we know

that we’ve seen a formula for

antidifferentiating a single power of x. I mean we know how

to differentiate a single power of x, and

so to do an antiderivative is just the inverse process. And we also know that when you

have the derivative of a sum, it’s the sum of derivatives. And so consequently, if you have

the antiderivative of a sum, it’s just the sum of

the antiderivatives. So we can write this out

into its constituent parts. So it’s the antiderivative

of x to the sixth dx plus– now of course you

don’t have to do this. You could probably proceed

just from this step onwards, or, but I don’t see any harm

in actually splitting it up myself. So antiderivative

of 2x*dx minus, OK, x to the minus 4/3 dx plus

minus 3 x to the minus 2 dx. So I’ve just split it up

into a bunch of pieces. I guess this one I sort of

pulled the minus sign out and this one I didn’t. But you know, whatever. Either way. OK so now we just

need to remember our formulas for taking the

antiderivative of a power of x. So in order to that, so

when you take a derivative, the power goes down by one. So if you take an

antiderivative to the power will always go up by one. So in this case

you get, so you’re going to get x to the seventh. And now when you differentiate

x to the seventh, a 7 comes down in front, right? You get 7 x to the sixth. So in order to get

just x to the sixth, we have to also divide

by that 7 there. So x to the sixth,

the antiderivative is x to the seventh over 7. 2x, so that’s going to give

us plus 2 x squared over 2. Or if you like, you could just

recognize right away that 2x is the derivative of x squared. Minus– OK now we’ve

got minus powers. Rather, negative powers, so

that always is a little trickier to keep track of. So again, the same

thing is true though. You have to, you add

one to the exponent. The exponent goes up by one

when you take an antiderivative. It goes down by one when

you take a derivative. So when you add 1 to minus

4/3 you get minus 1/3. So we have x to the minus 1/3. And now I have to

divide by minus 1/3. When I take a derivative here,

we get– of x to the minus 1/3, I get minus 1/3 x

to the minus 4/3. So I need to divide

by that minus 1/3. OK. And finally here so

minus 3 x to the minus 2. So OK, so just like

this first one, you might recognize

that right off as the derivative

of x to the minus 3. So this is plus– Oh! Ha ha! You could do that if you were

completely confused like me. So right, so x to the minus

2, it increased by one. Increases by 1. So when it increases by 1,

you get minus 1 not minus 3. Oh! OK, good. So this is minus 3 times x

to the minus 1 over minus 1. OK. That’s much better. And if you like,

right, so, OK, so we could– any constant

we add to this, it’ll still be an

antiderivative. And now we can do a

little bit of arithmetic to arrange this into

nicer forms if you wanted. So you could

rewrite this as say, x to the seventh

over 7 plus x squared plus 3 x to the minus 1/3

plus 3 x to the minus 1 plus a constant. Now, suppose you

got here and suppose that you did the same

mistake that I just made. And you had accidentally

thought that this was going to be a minus 1/3

power instead of a minus first power. So how would you,

is there any way that you can prevent

yourself making that mistake? Well there actually is. So one nice thing

about antiderivatives is that it’s really

easy to check your work. After you’ve computed an

antiderivative, or something that you think is

antiderivative, you can always go back and take

the derivative of the thing that you’ve computed

and check that it’s equal to what you started with. So if you, if

you’re ever worried that you made a mistake

computing an antiderivative, one thing you can always

do is take a derivative of what you’ve got at the end. So if we take a

derivative here we get x to the sixth plus 2x minus

x to the minus 4/3 minus 3 x to the minus 2. OK? So that was just using our

rule for powers one by one. And OK, so you say that

out loud or write it down and then you just check. Right? So I said that, so that’s

exactly the same thing we’ve got right here. Yeah? So x to the sixth plus 2x

minus x to the minus 4/3 minus 3x to the minus 2. So one of the nicest things

about antiderivatives, they can be difficult to

figure out in the first place, but after you’ve got

something that you think is antiderivative it’s very

easy to go back and check whether you did it correctly

by taking the derivative and making sure that it

matches the thing that you were trying to antidifferentiate

at the beginning. So that’s that.

## 12 Comments

## Juniper Leigh

Yay!!! Thank you…

## epic311714

great video. thanks

## T S

you saved my ass

## Josyli Mabansag

Thanks. 🙂

## imakeadifference2011

Thanks!

## Priscila Briones

such a good instructor

## Jasmine Kadavil

Excellent teaching and stress-relieving! Thanks 🙂

## Kitty Bobcat

@alex Joelsson subtracting fractions… refresh yourself

## Ytremz

Why do you call this computation 'Anti-Derivative'… in the UK we call the Integrals…

## Anwar Khan

Excellent video!

## anonymous clark

couple of minutes means few seconds ,lolx

## Ernesto Hendrix

he looks like jay weingarten