Fundamental Theorem of Calculus

You'd think math guys would have a better grip on the whole singular/plural thing. The Fundamental "Theorem" of Calculus is really two theorems that are typically stated together because one isn't much use without the other. Anyway, it's the type of thing that might come up on the Q. Since I'm not even very good at remembering other people's results, much less proofs of such results, I decided the best thing to do would be to write my own proof. It's a little hand-wavy, in that I tend to just claim that things converge rather than busting out the obligatory epsilons and deltas, but it illustrates the point and I'm pretty sure such rigor won't be required on a timed exam.

I also needed to minimize the reliance on other results, so I got it down to where I don't use much other than the completeness of the reals (to get convergence) and the mean value theorem (to get equality on the derivatives). FWIW, here it is. There are much better proofs out there, but, because I derived this one, I'll be able to remember it.

Part i) If f is continuous and differentiable on the interval [a,b] and f' is Riemann integrable on [a,b], then the integral of f' from a to b is f(b) - f(a).

Proof: Let P be a partition of [a,b] and T={t_i} be a set of points such that t_i sits in the component i of P. The Riemann-Stieltjes sum:



Is bounded by the Upper and Lower Riemann sums on P. Since we get to choose the points in T, we can use the mean value theorem to select each t_i such that



Thus,



since all the middle points cancel out.

f' integrable implies that S converges to the integral as the norm of P goes to zero. Thus:



Part ii) If f is Riemann integrable on [a,b] and F(x) is the integral from a to x and F(a)=0, then F is continuous on [a,b] and F'=f where ever f is continuous.

f integrable on a closed interval implies f is bounded on that interval. Let M ≥ |f(x)|. Then,



Thus, F is continuous at x from the right. A symmetrical argument shows it is continuous from the left.

If f is continuous at x, then by the mean value theorem for integrals, there exists δ>0 and x₀ in (x,x+δ) such that



Taking the limit as δ goes to zero gives



Again, a symmetrical argument shows the same result from the left. Therefore, F'(x)=f(x).

Q.E.D.

That's latin for done, y'all.