The Unapologetic Mathematician

Mathematics for the interested outsider

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