If $f \colon [a,b] \rightarrow \mathbb{R}$, then show that there exists $c \in [a,b]$ such that $\int_a^c f = \int_c^b f$ If $f \colon [a,b]\rightarrow \mathbb{R}$, then show that there exists $c \in [a,b]$ such that $c \in [a,b]$ such that $\int_a^c f = \int_c^b f$.
I think that I have to proceed with the Mean Value Theorem of Integrals...
 A: If $f(x)$ is a continuous function in $[a,b]$, we may define a function $F(x)$ such that
$F(x) = \int_a^x f(y)\;dy$

$F(x)$ will be continuous in $[a,b]$ and differenciable in $(a,b)$:
$F'(x) = f(x)$
$F(a) = 0 \;\; ; \;\; F(b) = \alpha$
(Fundamental Theorem of Calculus)

By splitting the integrals, the problem becomes: show that $\,\exists \, c\in(a,b)$ such that
$F(c) = F(b) - F(c) \Leftrightarrow F(c) = \frac{1}{2}F(b)$
Since $F(c) \in[0,\alpha]$, the existence of $c$ can be directly proved by the
Intermediate Value Theorem
A: For $a\leq x \leq b$, let
$$F(x) = \int_a^x f(t)\, dt$$
We want to find $c\in [a,b]$ such that $F(x)=\frac 1 2 F(b)$. $F(0)=0$, so either $0 \leq \frac 1 2F(b) \leq F(b)$ (in the case that $F(b)\geq 0$) or $0 \geq \frac 1 2F(b) \geq F(b)$ (in the face that $F(b)\leq 0$). In either case, since $F$ is continuous, the intermediate value theorem guarantees the existence of $c\in [a,b]$ such that $F(c)=\frac 1 2 F(b)$. It follows that
$$\frac{1}{2}F(b)=\int_a^c f(t)\, dt = \int_c^b f(t)\, dt$$
