Definite Integral by Substitution | MIT 18.01SC Single Variable Calculus, Fall 2010

Uploaded by MIT on 07.01.2011


Welcome back to recitation.
Today we're going to do another example
of a definite integral.
This time, it's going to be one that you're going to need
to do some substitution for.
You're not going to know the antiderivative right off the
top of your head, I think.
So, I've got it on the board behind me.
So it's compute the integral from minus 2 to 2 of the
function x squared cosine of the quantity x
cubed over 8 dx.
So, this is a kind of weird a integrand.
I roughly, you know, very, very rough graph of kind of
what it looks like.
So this is sort of the area of this region.
It's a positive function over this interval.
So it's the area of this region that I'm
asking you to compute.
So, all right, so definite integral.
Why don't you pause the video, take a couple minutes, work
this out yourself, come back, we can work it out together.

All right, welcome back.
So, from looking at this integrand here, it seems to me
that I don't know immediately off the top of my head what
its antiderivative is.
So, I can't just sort of apply the fundamental theorem of
calculus directly without a little bit of work.
So there are two different ways that I can go about
computing this then, about computing
this definite integral.
One is, that I can forget for the minute that it's a
definite integral and compute the antiderivative and then
use the fundamental theorem of calculus.
So, let's try that way first and then we'll do it a second
way as well.
So, method one is to compute the antiderivative.
So we compute the antiderivative of x squared
cosine of x cubed over 8 dx.
So here, it's not a definite integral anymore.
Now I'm just looking at the antiderivative.
And so, OK, so either by intelligent guessing or that--
I forget if that's what Professor Jerison called it--
or by just a substitution method.
So, a good thing to substitute here, is we see inside this
function we have this x cubed part and then outside we have
an x squared part.
So x squared dx, that's closely related to the
differential of x cubed.
So we might try the substitution.
We might try the substitution u equals--
well we can maybe get rid of all of that at once-- so make
u equals x cubed over 8.
So in that case, du is equal to 3/8 x squared dx.
And so this integral is equal to the integral of--
well so, x squared dx, we have that.
So we need to multiply and divide by 3/8.
So this is, 8/3 cosine of u du.
All right, and so now, OK, so the 8/3,
that's just a constant.
That pulls out in front.
That's not hard.
So then the antiderivative of cosine u du, that's
something we know.
So this is 8/3 of minus sine--
sorry not minus; I was doing the derivative instead of the
Just 8/3 sine u.
And good.
So that's the anti plus the constant, which we don't need
because we're going to plug this in the fundamental
theorem of calculus in a minute.
So it's 8/3 sine u.

And, OK, so, but that's not good.
I need it back in terms of my x's.
And so I have to back substitute.
So this is 8/3 sine of x cubed over 8.
All right, now I've got an antiderivative of this
So now I can take this antiderivative and apply
fundamental theorem of calculus, Right?
So, the integral from minus 2 to 2 of x squared cosine of x
cubed over 8 dx is equal to--
well it's equal to this antiderivative--
8/3 sine x cubed over 8 taken between x equals minus 2
and x equals 2.
So this is equal to 8/3 times sine of 1 minus 8/3 times sine
of minus 1.
OK, and sine of minus x is equal to minus sine x.
So we could rewrite this as 16/3 times sine of 1.
So that's the answer.
That's the value of this integral.
And we did this first method by first computing the
antiderivative and then using it with the fundamental
theorem of calculus.
The second way we can do it is the one that we just learned
more recently in lecture.
Which is to use substitution directly in
the definite integral.
So let me come over here and do that.
So, method two.

So now, when we do it this way, mostly
this looks the same.
So we start off, we don't go back to antiderivative.
We start off with the definite integral.
So that's one difference.

We can use the exact same substitution.
So we can use exactly the substitution that we used
before, which was u equals x cubed over 8.
That doesn't change.

And we still have the same du is equal to 3x squared over 8,
but what does change-- sorry, dx.
You need a differential always to be equal to a differential.
So what does change is that this--
I need to change the bounds.
If I'm keeping it as definite integrals all the time, so I'm
changing here from an integral with respect to x to an
integral with respect to u.
And so that means that when I write down the second integral
with respect to u, the bounds that I write down have to be
the bounds for u.
Not for x.
So when x is equal to minus 2, I need to know what the
corresponding value of u is.
So in this case, when x is equal to minus 2, we see that
the lower bound here--
when x is equal to minus 2--
the lower bound on u becomes u equals, well, minus
2 cubed over 8.
That's minus 1.
And when x is equal to 2, the upper bound, u becomes
2 cubed over 8.
So u becomes 1.
So we have this extra step when we do it this way, of
changing the bounds.
OK, so I do that.
So now I get the integral.
So I'm going to write u equals minus 1 here just to remind
myself that I made this change to 1.
OK, and now the inside transforms exactly the same.
So it's 8/3 times cosine u du.
OK, and now again, this is something for which I already
know the antiderivative.
So I apply the fundamental theorem of calculus here.
So I get the integral is equal to 8/3 sine of u, where u goes
between minus 1 and 1.
So the other thing that's changed here is that now I
don't have to change this back in terms of x's.
I have the values that I'm going to plug in here already
in terms of u.
So I don't have to switch back in terms of x's.
And so I just plug these values in and take the
So this is equal to 8/3 sine of 1 minus 8/3 sine of minus
1, which as we said before is 16/3 times sine of 1.
OK, so we have these two different methods.
They're very, very, very, similar in how you apply them.
The key differences are, that when you do it, the
substitution in terms of definite integrals, you have
to change your bounds of integration.
And what you get for changing your bounds of integration is
at the end you don't have to switch back from u's into x's.
Whereas, if you do it by first computing the antiderivative,
well, you don't have any bounds of integration when you
compute the antiderivative so you don't have to change them.
But once you computed the antiderivative, then you have
to go and make your back substitution.
Put everything back in terms of x's before going on to
compute the definite integral that you started with.
So the those are the two different ways you can do
substitution in definite integrals and
I'll end with that.