Show that there exists a $c : f(c) = g(c)$ I'll try to present a solution for this problem, and I hope I can receive feedback on what went wrong, if something went wrong of course.


Let $f, g : [a, b] \to \Bbb R$ be continuous functions and $\int_{a}^{b} f(x) dx = \int_{a}^{b} g(x) dx$. Show that there exists $c \in [a, b]$ such that $f(c) = g(c)$.


Solution
Let's define $h(x) = \int_{a}^{x}f(x)dx-\int_{a}^{x}g(x)dx$
$h(x)$ is continous, since $f(x)$ and $g(x)$ is continous. I hope this argument is correct.
We see $h(a) = h(b) = 0$.
Applying Rolle's Theorem, we get that $\exists \xi \in (a,b) : h'(\xi) = 0$
In other terms,
$f(\xi) = g(\xi)$
$\square$

Thanks!

Instead of claiming $h$ to be continuous, you need that $h$ is differentiable, in order to apply Rolle's Theorem. The differentiability follows since $f$ and $g$ are continuous, so then the Fundamental Theorem of Calculus tells you that $h$ is differentiable.
Apart from that, the solution is correct.

Here's another method: Assume for the sake of contradiction that the statement is not true. Define $h := f - g$. By assumption, $h$ is never zero. Thus, $h$ is of one sign. WLOG, $h > 0$. But this means that $\int_a^b h > 0$, contradicting our assumption.
(The work here is gone into showing that $h > 0 \implies \int_a^b h > 0$.)
