Okay, in this video I want to show
you how to combine
the fundamental theorem of calculus
and the chain rule to get some
pretty strong results.
You sometimes get problems like
compute the derivative of
the integral from 3x to x^2 of
e^{-s^2} ds.
At a first glance, this looks horrible.
How are you ever supposed
to find that integral?
And the once you know how to find
that integral, how are supposed
to take its derivative?
In fact, we know that
the integral does make sense, we
know that there is an antiderivate
of e^{-s^2}, we just don't know
the formula for it.
So let F(x) be the antiderivative
of f(x), which is e^{-x^2}.
So we're trying to take the derivative
of the integral from 3x to x^2
of f(s) ds.
And that's the derivative, well we
use the second fundamental theorem
of calculus and say to do this integral
from 3x to x^2, we just take F(x^2)-F(3x).
Now, we don't happen to know a
formula for F, but that's okay
because now we apply the chain rule.
The way you take the derivative of
F(x^2) is you take F'(x^2)
times the derivative of x^2.
And the derivative of F(3x)
is F'(3x) times three,
and now here comes the magic.
F is an antiderivative,
so the derivative of F is f,
so you just plug it in.
And since f(x)=e^{-x^2}, this gives
you e^{-x^4}*2x,
minus e^{-9x^2}*3.
And there's your answer.
In general, if you're ever given
problems that look like
find the derivative of the integral
from, let's call it g(x), to h(x)
of f(s) ds, you say oh, that's the
derivative of F(h(x))-F(g(x)),
and you don't worry about the fact
that you don't know a formula for F
because you're going to take its
derivative, and you do have a formula
for the derivative of F, it's f.
So this is always going to give you
F'(h(x)) times h'(x) minus F'(g(x))
times g'(x) and that's going to be f(h(x))
times h'(x) minus f'(g(x)) times g'(x).
So you often run into these kinds
of problems, and by combining
the chain rule with the second fundamental
theorem of calculus, we're not
actually using the first fundamental
theorem, we're using
the second fundamental theorem,
we get the answer.
In fact, this is a backdoor way of
seeing why the first
fundamental theorem is true.
If you believe that every function
has an antiderivative, and for that,
you really do need to go through
the proof for the first
fundamental theorem, but if you believe
that antiderivatives exist,
then the derivative of the integral
from a to x of f(s) ds, well
that's the derivative of F(x)-F(a).
The second fundamental theorem of
calculus says you find an antiderivative
and you plug in at the end points
and that's how you do an integral
from one spot to another spot,
and that's F'(x)-0 because this is
a constant, it doesn't depend on x,
so its derivative is 0, and that's f(x).
So that's another way of understanding
the first fundamental theorem of calculus.
Once you have the second
fundamental theorem of calculus,
the first fundamental theorem of
calculus comes almost for free.
