As Davide suggested, put $F(x)=\displaystyle\int_0^x f(t)dt$. Then $F(0)=\displaystyle\int_0^0 f(t)dt=0$ and $F(1)=\displaystyle\int_0^1 f(t)dt=2a$ since $a=\displaystyle\frac{1}{2}\int_0^1 f(t)dt$, which shows that $F(0)=0\leq a\leq 2a=F(1)$. Note that $F$ is continuous since $f$ is integrable, by Intermediate value theorem, there exists $c\in [0,1]$ such that $\int_0^c f(t)dt=F(c)=a.$
To prove that $c$ is unique, suppose by contradiction that there exists c'\neq c such that F(c')=a. By definiton of $F$, we have \int_0^{c'} f(t)dt=F(c')=a=F(c)=\int_0^{c} f(t)dt. Without loss of generality, we assume c. Hence, we have \displaystyle\int_c^{c'} f(t)dt=0. By this is a contradiction, because by assumption $f(x)>M$ we have \displaystyle\int_c^{c'} f(t)dt\geq M(c'-c)>0.
Note: In my previous answer, I made a mistake by assuming that $f$ is continuous. Thank you for Didier pointing out the mistake.
Note added: Proof of $F$ being continuous: $|F(x)-F(y)|=\left|\int_0^{x} f(t)dt-\int_0^{y} f(t)dt\right|\leq\int_{[x,y]}|f(t)|dt \rightarrow 0\mbox{ as }x\rightarrow y$ since $f$ is integrable.