Is the integral squared equal to two times the integral from $a$ to $b$ and from $x$ to $b$? I have to prove that 
$$2\int_a^b\int_x^bf(x)f(y)\,dx\,dy = \left(\int_a^bf(x)\,dx\right)^2  $$
where $f$ is continuous in $[a,b].$
I tried to separate the integrals in a way that I get that 
\begin{align*}
\left(\int_a^bf(x)\,dx\right)^2 &= \left(\int_a^bf(x)\,dx\right) \left(\int_a^bf(y)\,dy\right)\\
&= \left(\int_a^bf(x)\,dx\right)\left(\int_a^xf(y)\,dy+\int_x^bf(y)\,dy\right).
\end{align*}
But I don't understand how am I supposed to prove that
$$\int_a^xf(y)\,dy=\int_x^bf(y)\,dy$$
becuase that expression depends on the $x$ that I take.  It seems that I need a certain symmetry in the function to accomplish something similar, where I can take another $x_{o}$ in $[a,b]$ and show that
$$\int_a^{x_{o}}f(y)\,dy=\int_{x_{o}}^bf(y)\,dy.$$
There something that I am not seeing, any hint or idea would be helpful.

Let's have $\displaystyle F(x)=\int_a^xf(t)dt$
Since $f$ is continuous on $[a,b]$ then $F$ is $C^1$ and $F'(x)=f(x)$.
$\displaystyle I_1=\bigg(\int_a^b f(x)dx\bigg)^2=F(b)^2$
$\begin{array}{l}
\displaystyle I_2=2\int_a^b\bigg(\int_x^b f(x)f(y)dy\bigg)dx=2\int_a^bf(x)\big(F(b)-F(x)\big)dx\\
\displaystyle\phantom{I_2}=2F(b)\int_a^bf(x)dx-\int_a^b 2F(x)F'(x)dx\\
\displaystyle\phantom{I_2}=2F(b)^2-\bigg[F(x)^2\bigg]_a^b=2F(b)^2-\bigg[F(b)^2-0\bigg]=F(b)^2\\
\phantom{I_2}=I_1\end{array}$
Note: do we used the continuity of $f$ or is it superfluous for the proof ?
