There exists $c\in [a,b]$ such that $\int_a^c f(t)dt = \int_c^b f(t)dt$ If $f:[a,b]\longrightarrow \mathbf{R}$ is integrable prove that there is $c\in[a,b]$ such that $\int_a^c f(t)dt = \int_c^b f(t)dt$.
I set $g(x)=\int_a^x f(t)dt$ but I don't know how I must continue.

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
