Problem: Let $f$ be a function such that the graph of $f$ is a semicircle $S$ with end points $(a,0)$ and $(b,0)$, where $a < b$. The improper integral $\int_{a}^{b} f(x) f'(x) dx$ is
(A) necessarily zero
(B) possibly zero but not necessarily zero
(C) necessarily nonexistent
(D) possibly nonexistent but not necessarily
(E) none of the above
This is a practice problem for timed test, so I am looking for the quickest solution. Here is what I came up with. Right from the beginning it is rather hand-wavy. It suffices to transform the circle into the unit circle centered at $(0,0)$. In that case, the integral becomes $\int_{-1}^{1} f(x) f'(x) dx$, where $f(-1)=f(1)$. Since $\frac{1}{2} \frac{d}{dx}f(x)^2 = f(x) f'(x)$, the integral becomes
$$\frac{1}{2} \int_{-1}^{1} dx f(x)^2 dx = \frac{1}{2} [f(1)^2 - f(-1)^2] = 0,$$
which means that (A) is the correct choice.
My question is, how do I make this idea of transforming the circle more rigorous? I have never actually done anything like this before, but I could see geometrically that it would work. How do I make this step more rigorous?
With the substitution $u = f(x)$, $du = f'(x) \, dx$, we get $$\int f(x) f'(x) \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{(f(x))^2}{2} + C.$$ Since $f(a) = f(b) = 0$, the result immediately follows.
It is also worth noting that we can explicitly compute $$f(x) = \pm \sqrt{((b-a)/2)^2 - (x-(a+b)/2)^2}, \\ f'(x) = \frac{-(x-(a+b)/2)}{f(x)},$$ so that $$f(x) f'(x) = \frac{a+b}{2} - x, \quad a < x < b.$$ But it is clear from the previous calculation that $f$ need not be explicitly stated: as long as $f(a)^2 = f(b)^2$ and certain regularity conditions hold (what are these?), then the integral will be zero.