The Unapologetic Mathematician

Mathematics for the interested outsider

The Fundamental Theorem of Calculus (all together now)

So we’ve seen two sides of the FToC: the first part, which says that given a continuous function f:\left[a,b\right]\rightarrow\mathbb{R} we can integrate and differentiate to get our function back:


and the second part, which says that given a differentiable function F:\left[a,b\right]\rightarrow\mathbb{R} whose derivative is the continuous function f, we can integrate to get (part of) our function back again:

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

Now, we proved these two sides in very different ways, but it turns out that we can get from one to the other.

Let’s assume the first part holds. Then we take the function F and define f(x)=F'(x) as its derivative. The first part of the theorem tells us that we know a function whose derivative is f: the function defined by G(x)=\int_a^xf(t)dt. And we know that any two functions with the same derivative must differ by a constant! That is, there is some real number C with F(x)=G(x)+C. Using this to evaluate F(b)-F(a) we find:

\displaystyle F(b)-F(a)=(G(b)+C)-(G(a)+C)=G(b)-G(a)=

Which gives us the second part of the theorem.

On the other hand, what if we assume the second part of the theorem holds? Then we start with a continuous function f:\left[a,b\right]\rightarrow\mathbb{R}. Given x\in\left[a,b\right], the function is continuous on the subinterval \left[a,x\right], and so the second part of the FToC says that F(x)-F(a)=\int_a^xf(t)dt. That is, the integral in the first part of the FToC differs by a constant (F(a)) from the function F we assumed to be an antiderivative of f. Thus it must itself be an antiderivative of f.

So each half of the Fundamental Theorem implies the other, and we can prove either one first before immediately deriving the other.

February 15, 2008 Posted by | Analysis, Calculus | 125 Comments



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