Show that exists $x \in [a,b]$ such that $\int_a^x f(t)dt = \int_x^b f(t)dt$ Suppose that $f$ is an admissible function (bounded and with the set of discontinuity with volume zero) in $[a,b]$. Show that exists a number $x \in [a,b]$ such that $$ \int_a^x f(t)dt = \int_x^b f(t)dt$$ **My attempt:** My guess is that I must use the fundamental theorem of calculus. If such $x$ exists, it must satify $$ F(x) = \frac{F(b) + F(a)}{2}$$ for $F' = f$ But I couldn't find such $x$. Any sugestion? Thanks in advance!

If you want to use intermediate value theorem(IVT), define $$G(x) = \int_a^x f(u) du - \int_x^b f(u) du$$

Observe that $G(a) = -G(b)$. Thus either they are both zero in which case you are done, or if $G(b) = \alpha > 0$, then $G(a) = -\alpha < 0$ and so by IVT, there exists $x$ such that $G(x)=0$.

This doesn't require fundamental theorem of calculus.