Prove that $f'(x)g(x) = 0$ for all $x \in [a,b]$ Let $f,g$ be continuously differentiable functions on $[a,b]$, suppose $g$ is nonnegative on $[a,b]$ and $f$ is increasing on $[a,b]$. Suppose further that $g(a) = 0$ and $g(b) = 0$, and that
$$\int_a^bf(x)g'(x)dx = 0$$
Prove that,
 $$f'(x)g(x) = 0$$ for all $x \in [a,b]$.
 A: By integration by parts, $\int f'(x) g(x) dx = f(x) g(x) - \int f(x) g'(x) dx$.
So $0 = \int_a^b f'(x) g(x) dx = f(a) g(a) - f(b)g(b) - \int_a^b f(x) g'(x) dx$
Since $g(a) = g(b) = 0$, it's clear that the first two terms are $0$.  The third term in the expression, we are given, is also $0$: $\int_a^b f(x) g'(x) dx = 0$.
We have $\int_a^b f'(x) g(x) dx = 0$, and $f'(x) g(x) \ge 0$ and is continuous, so we might conclude here that $f'(x) g(x) = 0$ for all $x \in [a,b]$.

If we cannot use the theorem that a nonnegative continuous function whose integral over some interval is identically $0$ over that integral, then we may proceed as follows:
By assumptions, $g(x) \ge 0$ and $f'(x) \ge 0$.  Suppose $f'(c) > 0$ and $g(c) > 0$ for some fixed point $c \in ( a,b)$.  By continuity, $f'(x) > 0$ and $g(x) > 0$ for some neighborhood $(c-\epsilon,c+\epsilon)$.  So $$\int_{c-\epsilon}^{c+\epsilon} f'(x) g(x) dx > 0$$
By nonnegativity of the product $f'(x) g(x)$, it follows that $$\int_a^{c-\epsilon} f'(x) g(x) dx , \int_{c+\epsilon}^b f'(x) g(x) dx \ge 0$$
So $\displaystyle \int_a^b f'(x) g(x) dx = \int_a^{c-\epsilon} f'(x) g(x) dx + \int_{c-\epsilon}^{c+\epsilon} f'(x) g(x) dx + \int_{c+\epsilon}^b f'(x) g(x) dx > 0$
