Computing Antiderivatives | MIT 18.01SC Single Variable Calculus, Fall 2010
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Computing Antiderivatives | MIT 18.01SC Single Variable Calculus, Fall 2010


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.

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