WEBVTT
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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 antiderivative
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of this big fraction.
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So on top it's got x to
the eighth plus 2x cubed
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minus x to the 2/3 minus 3,
that whole thing over x squared.
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So just a quick
linguistic note about why
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I said an antiderivative
instead of the antiderivative,
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and then I'll 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 any one
as an 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
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is equal to this function.
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So why don't you take a
couple minutes, work this out,
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come 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
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is that I've written this
in a sort of a silly form.
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And it's probably
a lot easier to get
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a feel for what this
function is if you
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break this fraction apart
into its several pieces.
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So for example, x to the
eighth over x squared
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is just x-- So,
well OK, so let me,
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this antiderivative
that I'm interested in,
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antiderivative of x to the
eighth plus 2x cubed minus x
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to the 2/3 minus 3,
over x squared dx.
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So I've written
this in a silly form
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and you can get
it in a nicer form
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if you just realize
that, you know,
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this is just a
sum of powers of x
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that I've put over this
silly 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
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of well, x to the
eighth over x squared.
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That's x to the sixth.
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And 2 x cubed over
x squared is just x.
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So I have x-- sorry,
it's 2x-- 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,
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so we could write that as minus
3 over x squared, or maybe it's
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a little more convenient
to write it as minus 3
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x 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
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a formula for
antidifferentiating
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a single power of x.
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I mean we know how
to differentiate
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a single power of x, and
so to do an antiderivative
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is just the 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 sum,
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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
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plus-- 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 splitting it
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up myself.
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So antiderivative
of 2x*dx minus, OK,
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x to the minus 4/3 dx plus
minus 3 x 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
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and 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
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our formulas for taking the
antiderivative of a power of x.
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So in order to that, so
when you take a derivative,
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the 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
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going to get x to the seventh.
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And now when you differentiate
x to the seventh,
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a 7 comes down in front, right?
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You get 7 x to the sixth.
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So in order to get
just x to the sixth,
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we have to also divide
by that 7 there.
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So x to the sixth,
the antiderivative
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is x to the seventh over 7.
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2x, so that's going to give
us plus 2 x 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-- OK now we've
got minus powers.
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Rather, negative powers, so
that always is a little trickier
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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 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 3 x to the minus 2.
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So OK, so just like
this first one,
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you might recognize
that right off
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as the derivative
of x to the minus 3.
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So this is plus-- 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
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we could-- any constant
we add to this,
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it'll still be an
antiderivative.
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And now we can do a
little bit of arithmetic
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to arrange this into
nicer forms if you wanted.
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So you could
rewrite this as say,
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x to the seventh
over 7 plus x squared
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plus 3 x to the minus 1/3
plus 3 x to the minus 1
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plus a constant.
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Now, suppose you
got here and suppose
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that you did the same
mistake that I just made.
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And you had accidentally
thought that this
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was going to be a minus 1/3
power instead of a minus
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first power.
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So how would you,
is there any way
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that you can prevent
yourself making that mistake?
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Well there actually is.
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So one nice thing
about antiderivatives
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is that it's 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,
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you can always go back and take
the derivative of the thing
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that you've computed
and check that it's
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equal to what you started with.
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So if you, if
you're ever worried
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that you made a mistake
computing an antiderivative,
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one thing you can always
do is take a derivative
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of what you've got at the end.
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So if we take a
derivative here we
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get x to the sixth plus 2x minus
x to the minus 4/3 minus 3 x
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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
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and then you just check.
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Right?
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So I said that, so that's
exactly the same thing
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we've 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
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minus 3x to the minus 2.
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So one of the nicest things
about antiderivatives,
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they can be difficult to
figure out in the first place,
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but after you've got
something that you think
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is antiderivative it's very
easy to go back and check
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whether you did it correctly
by taking the derivative
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and making sure that it
matches the thing that you
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were trying to antidifferentiate
at the beginning.
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So that's that.