Intermediate value theorem for a definite integral function

Question

If $f:[a,b]\rightarrow\mathbb{R}$ is continuous on $[a,b]$, show that $ \exists c\in[a,b]$ so that $$\int_{c-a}^{b-c}f(x) \, dx=0$$ I think it can be solved using the intermediate value theorem but I can't find a suitable function.

Answer 1

If it makes sense, equivalently, $c-a, c-b\in [a,b]$ (for example if $[a,b]=[0,1]$) for every $c\in [a,b]$ define $g(x)=\int_{c-a}^{b-c}f(x) \, dx$, $g(a)=-g(b)$

Take $c\in[50,51]$ for example, $c-50, 51-c\in [0,1]$.

Answer 2

As stated, is true even without the continuity hypothesis. Take $b - c = c - a$, i.e., $c = (a+b)/2$ the midpoint of the interval.