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

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 $x\not\in (-1,1)$.
