Is it always possible to choose $x \in (a, b)$ s.t. $\int_a^{x}f=\int_x^{b}f$

Question

I am working out a homework problem about Riemann Integrals and the question is as follows:

Suppose that $f$ is integrable on $[a, b]$, then $\exists \ x \in [a, b] s.t. \int_a^{x}f=\int_x^{b}f$. Is it always possible to choose $x$ to be in $(a, b)$?

I have managed to prove the first part and now I am attempting the second part of the question.

This is my reasoning:

Let $f$ be a function such that choosing $x=a$ means $\int_a^{a}f=\int_a^{b}f$.

Now $\int_a^{a}f=0 \implies \int_a^{b}f=0$ and for this to be true, $f$ must be a function defined at only one point, ie $a=b$, which brings me to my question: is a function defined at a only one point Riemann Integrable and is the rest of my reasoning correct?

Answer 1

Yes, provided that $\int_a^b f\ne 0$.

For example, if $f(x)=x$, and $[a,b]=[-1,1]$, then $\int_{-1}^x t\,dt=\frac{1}{2}(x^2-1)$, while $\int_x^1 t\,dt=\frac{1}{2}(1-x^2)$ and $$ \int_{-1}^x t\,dt=\int_x^1 t\,dt\quad\Longrightarrow\quad x=\pm 1, $$ and hence $x\not\in (-1,1)$.