Calculus is one of the grandest achievements of human thought, explaining everything from planetary orbits to the optimal size of a city to the periodicity of a heartbeat. This brisk course covers the core ideas of single-variable Calculus with emphases on conceptual understanding and applications. The course is ideal for students beginning in the engineering, physical, and social sciences. Distinguishing features of the course include: 1) the introduction and use of Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus; 3) an emphasis on the conceptual over the computational; and 4) a clear, dynamic, unified approach. In this third part--part three of five--we cover integrating differential equations, techniques of integration, the fundamental theorem of integral calculus, and difficult integrals.
BONUS!
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From the course by University of Pennsylvania
Calculus: Single Variable Part 3 - Integration
ratings
From the lesson
The Fundamental Theorem of Integral Calculus
Indefinite integrals are just half the story: the other half concerns definite integrals, thought of as limits of sums. The all-important *FTIC* [Fundamental Theorem of Integral Calculus] provides a bridge between the definite and indefinite worlds, and permits the power of integration techniques to bear on applications of definite integrals.
Meet the Instructors
Robert GhristProfessor
Mathematics and Electrical & Systems Engineering
Welcome to calculus. I'm professor Ghrist.
We're about to begin lecture 26, bonus material.
In our main lesson, we covered the fundamental theorem of integral calculus.
We saw how it was used, and what it meant.
But we did not say why it is true. Let's sketch a proof, but how do we
proceed? Our goal is to show that the definite
integral of f from a to b is the indefinite integral evaluated at the
limits. We won't prove that directly rather, we
will prove a lemma, a preparatory step. The lemma is going to look very
different, but is really closely related. It states the following, the integral of
f of t dt, as t goes from a to x is what? Well, let's see, this is going to be a
function of x. If we differentiate that with respect to
x, what will we get? We will get f of x, the integrand
evaluated at x. Now this seems a little unusual since
you're differentiating a limit of a definite integral, but lets explore, see
if we can make sense of it. Lets make sure we're not completely crazy
and check that this works in a simple case.
Lets differentiate the integral as t goes from a to x of a constant dt.
Well, we can do the definite integral of a constant, c.
And that's just going to give us c times the upper limit x minus c times the lower
limit, a. If we differentiate that function of x,
what do we get? We get c, the constant which is the
integrand we began with. So, this is not a clearly crazy thing to
do. Well, let's see if we can prove this
lemma. Consider what the integrand f of t looks
like, we need to compute the definite integral from a to x.
And the claim is that the derivative of the definite integral is f evaluated at
x. Well, lets see, let's denote by capital
F, the definite integral as a function of x.
Then, if we want to differentiate that with respect to x, what should we do?
Let's look at what happens when we increase x by a small amount.
Let's call that h. Well, according to the definition, that
is the integral of f of t dt, as t goes from a to x plus h.
Now of course, we could write that as the integral as t goes from a to x, plus the
integral as t goes from x to x plus h. That comes from additivity of the
integral. And since by definition, the integral
from a to x is simply capital F at x, we see something that should start looking
like the definition of a derivative. Namely capital F at x plus h equals
capital F at x plus something. What is that something?
That something is going to have the derivative of capital F built into it.
Now, what we need to focus on is this interval from x to x plus h.
And here, I'm going to I have to make a little bit of a more restrictive
assumption about the integrand f namely, that not only is it continuous but it
also is reasonable. lets say has a a Taylor series associated
to it. Now, for this definite integral, lets
choose a partition with width h. What is the height?
Well, the height if we choose the left hand endpoint would be f of x.
The width is h now, this is not exactly what the definite integral is and it's
just an approximation, there's some leftover stuff that we haven't accounted
for. How can we estimate that?
Well, if the intagrand f has a a reasonable form to it, let's say it has a
well defined derivative in the sense of a Taylor series.
This, change in the height, this change in little f, is going to be big O of h.
That is, it shrinks to zero linearly as h goes to zero.
So, the worst thing that could happen, the estimate for what we left out here is
something that is big O of h times the width h.
Now, if we take big O of h times h, of course that's big O of h squared.
And now stepping back we see that we've computed the derivative of capital F of x
because capital F of x plus h is capital F of x plus something times h, plus
something in big O of h squared. What is that coefficient in front of the
h term? That first-order variation is, little f
evaluated at x. That is what we're trying to prove.
And now let's see what that does for us. The indefinite integral of f is what?
Well, from this lemma, we see that the definite integral of f of t dt as t goes
from a to x is an indefinite integral because it's derivative is little f of x.
Therefore, the general indefinite integral of f is, this definite integral
of f of t as t goes from a to x plus an arbitrary constant.
We now have an explicit form, the antiderivative in terms of a definite
integral. Now, what do we have to do?
Well, if we were to evaluate that indefinite integral as x goes from a to
b, what would we get? We would get the integral from a to x of
f of t dt plus some constant, c evaluated at b, minus the same thing evaluated at
a. Lets fill that in.
When we valuate at b, we get the integral of f of t dt.
As t goes from a to b, plus a constant. We subtract off the integral as t goes
from a to a of f of t dt plus a constant. I'm going to ignore the plus c, because
we first added and then subtracted. Now what do we notice?
Well, the integral from a to a is always zero, and so we're left with the integral
from a to b. But, notice that we're using t whereas as
we stated our theorem in terms of x. But of course, that doesn't matter a bit.
You can use any symbol you want for the definite integral.
And that is exactly what we were trying to show, the the indefinite integral
evaluated from a to b is the definite integral from a to b.
Well, that's nice as a proof, but one of the wonderful things about this proof is
that it gives us a new tool for computing some interesting derivatives.
Derivatives of definite integrals with functions in the limits.
Let's do a more complicated example. Consider the derivative with respect to x
evaluated at x equals 1 of the following function.
The integral of e to the t squared dt, as t goes from log of x to square root of x.
And this ostensibly is difficult. You can't find the antiderivative of e to
the t squared, easily if at all. But, we can still compute this
derivative. How do we do that?
Well, we've gotta be a bit careful. Our lemma only applies in the case where
the limits are from a to x. And here we have functions of x.
So, let's think for a moment. What we know, is that the derivative, the
integral from a to x of f of t dt is f of x.
What is going to help us, is to have the function of x at the upper limit.
So, lets rewrite this integral using additivity as the integral from 1 to
square root of x plus the integral from log of x to 1.
And now, using the orientation property, we can reverse the limits of the second
integral and subtract it. So, we're subtracting the integral from 1
to log of x. Now, 1 plays the role of a, the constant
at the lower limit. At the upper limits, we don't have x, we
have a function of x and so, what we're going to do is rethink our lemma in terms
of some function u of x, at the upper limit.
But we must be careful to use the chain rule to get the derivative with respect
to x of the integral from a to u of f of t dt is f of u du dx.
And now, we can proceed if we differentiate the integral from 1 to
square root of x of e to the t squared dt.
We get e to the square root of x squared, times the derivative of square root of x.
When we subtract off the second integral, that is e to the log squared of x, times
the derivative of log of x. This is the derivative, that
complicated-looking integral, we were asked to evaluate it at x equals 1.
And the rest is simply algebra. We take e to the x over 2 root x minus e
to the log squared x over x. Evaluate at x equals 1 and you can check
that we get one half e minus 1. You will find this method useful in
several problems. This is just one approach to the proof of
the fundamental theorem of integral calculus.
We made some [UNKNOWN] restrictive assumptions in using our Taylor series
understanding of what a derivative is. There are other proofs using more
sophisticated means, limits perhaps that will give you a more general result.
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