Prove that for any Riemann integrable $f$ on the interval $[a, b]$, there is a $c$ such that $\int_{a}^{c} f = \int_{c}^{b} f$

Question

This is one of the practice problems I am having a bit of trouble trying to figure out and prove.

Prove that for any Riemann integrable function $f$ on the interval $[a, b]$, there is a $c$ such that $\int_{a}^{c} f= \int_{c}^{b} f$.

I was thinking about like if $[a,b]$ is the entire interval, then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$. I am not sure how to proceed from here. Some help would be nice.

Answer 1

Let $g(x)=\int_x^bf-\int_a^xf$. Then, $g(a)=\int_a^bf$ and $g(b)=-\int_a^bf$. If $\int_a^bf=0$, you can use $c=a$. Else, you can use the Intermediate Value Theorem.

Answer 2

Let $F(x)=\int_a^x f(x)dx-\frac{1}{2}\int_a^b f(x)dx$. Since $F(b)=-F(a)$, and $F(x)$ is continuous, there is a point $x$ in the interval where $F(x)=0$, which is the desired point.