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\begin{document}
\title{The second fundamental theorem of calculus demystified}
\author{John Kerl}
\maketitle

We learned the second fundamental theorem of calculus:
if $f(x)$ is continuous and we let
$$
	F(x) = \int_0^x f(t) \,dt
$$
then
$$
	F'(x) = \frac{d}{dx} \int_0^x f(t)\,dt = f(x).
$$
We also have the chain rule:  if
$$
	\frac{d}{dx} F(g(x)) = \frac{d}{dx} \int_0^{g(x)} f(t)\,dt
	= f(g(x)) g'(x).
$$

\emph{What the heck does that mean}???

I'll address this question in three steps:
\begin{itemize}
\item[(1)]  What do these integral functions look like?
\item[(2)]  How do they relate to what we already know?
\item[(3)]  What does the chain rule have to do with anything?
\end{itemize}

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Let's start off with some things we can do easily.

\textbf{First example.}
What is
$$
	\int_0^{\pi/2} \cos(t)\,dt?
$$
Well, we use the (first) fundamental theorem of calculus, which says that
if (big if!) the integrand is the derivative of some other function $F(t)$,
then
$$
	\int_a^b f(t) dt = \int_a^b F'(t) dt = F(b) - F(a).
$$
And of course we have the shorthand notation
$$
	F(t) |_{t=a}^{t=b} = F(b) - F(a).
$$
I chose $f(t)=\cos(t)$ as an integrand because we all know what it's the
derivative of:  $F(t)=\sin(t)$.  (Generally, $\sin(t)+C$ for any constant
$C$.)  So
$$
	\int_0^{\pi/2} \cos(t)\,dt = \sin(t) |_{t=0}^{t=\pi/2}
	= \sin(\pi/2)-\sin(0) = 1.
$$
This makes sense because [xxx insert matlab figure here].

\textbf{Second example.}
OK, so let's do another one.  Suppose we want to let the right endpoint vary.
That shouldn't stop us, and it does not:
$$
	\int_0^x \cos(t) \,dt = \sin(t) |_{t=0}^{t=x} = \sin(x) - \sin(0) = \sin(x).
$$
This makes sense because [xxx insert matlab figure here].

We used the old function $\cos(x)$ to construct a new function $\sin(x)$.
Since this is a calculus class and we take derivatives of things, it's natural
to ask about the rate of change of this function.  We can do this two ways.
One way is to differentiate the new function:
$$
	\frac{d}{dx} \sin(x) = \cos(x).
$$
The other way is to remember how we constructed that new function, and use the
second fundamental theorem:
$$
	\frac{d}{dx} \int_0^x \cos(t)\,dt = \cos(x).
$$
Remember that the SFTC says that the derivative of the integral is found by
simply evaluating the integrand at the (moving) right end-point.

The key point:  Doing it the first way, we had to integrate and then
differentiate.  Doing it the second way, we just had to evaluate the integrand
at the right endpoint.

\textbf{Third example.}
That was perhaps unimpressive.  So what?  If I give you something really messy,
then it's non-trivial to find the integral:
$$
	\int_0^x \frac{e^{at}}{b+ce^{at}}\,dt
$$
However, if I asked you for the \emph{derivative} of this mess, then you're
in luck.  Instead of doing the hard work to find the integral, and then doing
more work to differentiate that, you can just use the second fundamental
theorem:
$$
	\frac{d}{dx} \int_0^x \frac{e^{at}}{b+ce^{at}}\,dt
	= \frac{e^{at}}{b+ce^{at}} |_{t=0}^{t=x}
	= \frac{e^{ax}}{b+ce^{ax}} - \frac{1}{b+c}.
$$

\textbf{Second example, part (b).}
OK, so let's do another one.  Suppose we want to let the right endpoint vary.
That shouldn't stop us, and it does not:
$$
	\int_0^x \cos(t) \,dt = \sin(t) |_{t=0}^{t=x} = \sin(x) - \sin(0) = \sin(x).
$$
This makes sense because [xxx insert matlab figure here].


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