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PROFESSOR: Hi.

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Welcome back to recitation.

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We've been talking in lecture
about antiderivatives.

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So I have here a problem
for you.

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Just an exercise about computing
an antiderivative.

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So the question is
to compute an

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antiderivative of this big fraction.

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So on top it's got x to the
eighth plus 2x cubed minus x

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to the 2/3 minus 3, that whole
thing over x squared.

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So just a quick linguistic
note about why I said an

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antiderivative instead of the
antiderivative and then I'll

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let you work on it a little.

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So an antiderivative.

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There are many functions whose
derivative is this function.

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Right?

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So they all differ from each
other by constants.

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So I would be happy
with and one as an

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answer to this question.

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That's why I chose the word
an antiderivative.

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So I'm looking for a function
whose derivative is equal to

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this function.

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So why don't you take a couple
minutes, work this out, come

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back and you can check your
answer against my work.

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OK, welcome back.

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So we were just talking about
this antiderivative here.

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So one thing you'll notice about
this function is that

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I've written this in a
sort of a silly form.

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And it's probably a lot easier
to get a feel for what this

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function is if you break this
fraction apart into its

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several pieces.

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So for example, x to
the eighth over x

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squared is just x--

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So, well OK, so let me, this
antiderivative that I'm

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interested in, antiderivative
of x to the eighth plus 2x

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cubed minus x to the 2/3 minus
3, over x squared dx.

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So I've written this in a silly
form and you can get it

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in a nicer form if you just
realize that, you know, this

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is just a sum of powers of x
that I've put over this silly

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common denominator.

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So our life will be a little
simpler if we write this out

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by splitting it up into the
separate fractions.

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So if I do that, this is just
equal to the antiderivative of

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well, x to the eighth
over x squared.

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That's x to the sixth.

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And 2x cubed over x
squared is just x.

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So I have x-- sorry 2x--

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plus 2x.

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Now, OK so x to the 2/3
over x squared.

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So that's x to the
2/3 minus 2.

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Which is x to the minus 4/3.

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And minus 3 over x squared, so
OK, so we could write that as

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minus 3 over x squared, or
maybe it's a little more

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convenient to write it as minus
3x to the minus 2 dx.

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So far I haven't really done
anything, you know.

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A little bit of algebra here.

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OK, but now we know that we've
seen a formula for

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antidifferentiating a
single power of x.

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I mean we know how to
differentiate a single power

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of x, and so to do an
antiderivative is just the

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inverse process.

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And we also know that when you
have the derivative of a sum,

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it's the sum of derivatives.

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And so consequently, if you have
the antiderivative of a

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sum, it's just the sum of
the antiderivatives.

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So we can write this out into
its constituent parts.

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So it's the antiderivative of
x to the sixth dx plus--

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now of course you don't
have to do this.

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You could probably proceed just
from this step onwards,

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or, but I don't see any
harm in actually

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splitting it up myself.

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So antiderivative of 2xdx minus,
OK, x to the minus 4/3

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dx plus minus 3x to
the minus 2 dx.

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So I've just split it up
into a bunch of pieces.

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I guess this one I sort of
pulled the minus sign out and

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this one I didn't.

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But you know, whatever.

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Either way.

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OK so now we just need to
remember our formulas for

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taking the antiderivative
of a power of x.

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So in order to that, so when
you take a derivative, the

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power goes down by one.

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So if you take an antiderivative
to the power

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will always go up by one.

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So in this case you get,
so you're going to

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get x to the seventh.

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And now when you differentiate
x to the seventh, a 7 comes

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down in front, right?

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You get 7x to the sixth.

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So in order to get just x to
the sixth, we have to also

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divide by that 7 there.

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So x to the sixth, the
antiderivative is x to the

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seventh over 7.

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2x, so that's going to give
us plus 2x squared over 2.

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Or if you like, you could just
recognize right away that 2x

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is the derivative
of x squared.

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Minus--

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OK now we've got minus powers.

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Rather negative powers, so
that always is a little

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trickier to keep track of.

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So again, the same thing
is true though.

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You have to, you add one
to the exponent.

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The exponent goes up by
one when you take an

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antiderivative.

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It goes down by one when
you take a derivative.

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So when you add 1 to minus
4/3 you get minus 1/3.

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So we have x to the minus 1/3.

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And now I have to divide
by minus 1/3.

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When I take a derivative here we
get, of x to the minus 1/3,

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I get minus 1/3 x to
the minus 4/3.

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So I need to divide
by that minus 1/3.

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OK.

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And finally here so minus
3x to the minus 2.

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So OK, so just like this first
one, you might recognize that

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right off as the derivative
of x to the minus 3.

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So this is plus--

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Oh!

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Ha ha!

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You could do that if you were
completely confused like me.

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So right, so x to the minus
2, it increased by one.

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Increases by 1.

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So when it increases by 1, you
get minus 1 not minus 3.

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Oh!

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OK, good.

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So this is minus 3 times x to
the minus 1 over minus 1.

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OK.

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That's much better.

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And if you like, right, so, OK,
so we could, any constant

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we add to this, it'll still
be an antiderivative.

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And now we can do a little bit
of arithmetic to arrange this

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into nicer forms
if you wanted.

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So you could rewrite this as
say, x to the seventh over 7

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

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1 plus a constant.

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Now, suppose you got here and
suppose that you did the same

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mistake that I just made.

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And you had accidentally thought
that this was going to

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be a minus 1/3 power instead
of a minus first power.

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So how would you, is there any
way that you can prevent

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yourself making that mistake?

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Well there actually is.

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So one nice thing about
antiderivatives is that it's

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really easy to check
your work.

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After you've computed an
antiderivative, or something

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that you think is
antiderivative, you can always

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go back and take the derivative
of the thing that

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you've computed and check
that it's equal to

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what you started with.

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So if you, if you're ever
worried that you made a

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mistake computing an
antiderivative, one thing you

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can always do is take a
derivative of what you've got

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at the end.

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So if we take a derivative of
here we get x to the sixth

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plus 2x minus x to the minus 4/3
minus 3x to the minus 2.

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OK?

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So that was just using our rule
for powers one by one.

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And OK, so you say that out loud
or write it down and then

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you just check.

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Right?

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So I said that, so that's
exactly the same thing we've

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got right here.

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Yeah?

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So x to the sixth plus 2x minus
x to the minus 4/3 minus

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3x to the minus 2.

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So one of the nicest things
about antiderivatives, they

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can be difficult to figure out
in the first place, but after

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you've got something that you
think is antiderivative it's

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very easy to go back and check
whether you did it correctly

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by taking the derivative and
making sure that it matches

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the thing that you
were trying to

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antidifferentiate at the beginning.

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So that's that.

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