# The Unapologetic Mathematician

## The Fundamental Theorem of Calculus II

And now we come to the second part of the FToC. This takes the first part and flips it around.

We again start with a continuous function $f:\left[a,b\right]\rightarrow\mathbb{R}$, but now we take any antiderivative $F$, so that $f(x)=F'(x)$. Then the FToC asserts that

$\displaystyle F(b)-F(a)=\int\limits_a^bf(x)dx$

Before we differentiated a function we got by integrating to get back where we started. Now we’re integrating a function we get by differentiating, and again get back where we started. Integration and differentiation are two sides of the same coin.

Let’s consider a partition of $\left[a,b\right]$ with points $a=x_0,x_1,...,x_{n-1},x_n=b$. Then we see that $F(b)-F(a)=F(x_n)-F(x_0)$. We can add and subtract the value of $F$ at each of the intermediate points to see that

$\displaystyle F(b)-F(a)=F(x_n)-F(x_{n-1})+F(x_{n-1})-...-F(x_1)+F(x_1)-F(x_0)$
$\displaystyle F(b)-F(a)=\sum\limits_{i=1}^nF(x_i)-F(x_{i-1})$

Now the Differential Mean Value Theorem tells us that there’s a point $c_i\in\left[x_{i-1},x_i\right]$ so that $F(x_i)-F(x_{i-1})=(x_i-x_{i-1})F'(c_i)$. And we assumed that $F'(c_i)=f(c_i)$, so we have

$\displaystyle F(b)-F(a)=\sum\limits_{i=1}^n(x_i-x_{i-1})f(c_i)$

But this is a Riemann sum for the partition we chose, using the points $c_i$ as the tags. Since every partition, no matter how fine, has such a Riemann sum, the integral must take this value, and the second part of the FToC holds.

February 14, 2008 Posted by | Analysis, Calculus | 18 Comments