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.
 A: 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 ?
A: Let be $\displaystyle F( x)$ the antiderivative of $\displaystyle f( x)$, therefore $\displaystyle F'( x) =f( x)$, the definite integral on interval $\displaystyle [ a,b]$ is $\displaystyle F( b) -F( a)$
\begin{gather*}
R_{1} =\left(\int ^{b}_{a} f( x) dx\right)^{2} =( F( b) -F( a))^{2} =\\
=F( b)^{2} -2F( b) F( a) +F( a)^{2}
\end{gather*}
And
\begin{align*}
R_{2} &=2\int ^{b}_{a}\left(\int ^{b}_{x} f( x) f( y) \ dy\right) dx\\
&=2\int ^{b}_{a} f( x)[ F( b) -F( x)] dx\\
&=2\left(\int ^{b}_{a} f( x) F( b) dx-\int ^{b}_{a} f( x) F( x) dx\right)\\
&=2\left( F( b)\int ^{b}_{a} f( x) dx-\int ^{b}_{a} F( x) F'( x) dx\right)\\
&=2\left( F( b)[ F( b) -F( a)] -\int ^{F( b)}_{F( a)} udu\right)\\
&=2\left( F( b)^{2} -F( b) F( a) -\left[\frac{u^{2}}{2}\right]^{F( b)}_{F( a)}\right)\\
&=2\left( F( b)^{2} -F( b) F( a) -\left[\frac{F( b)^{2}}{2} -\frac{F( a)^{2}}{2}\right]\right)\\
\\
&=2\left( F( b)^{2} -\frac{F( b)^{2}}{2} -F( b) F( a) +\frac{F( a)^{2}}{2}\right)\\
&=2\left( F( b)^{2}\left( 1-\frac{1}{2}\right) -F( b) F( a) +\frac{F( a)^{2}}{2}\right)\\
&=2\left(\frac{F( b)^{2}}{2} -F( b) F( a) +\frac{F( a)^{2}}{2}\right)\\
R_{2} &=F( b)^{2} -2F( b) F( a) +F( a)^{2} =R_{1}\\
\end{align*}
A: Here's how you can use the symmetry. Let $D_1$ be the triangle that you're integrating over on the left-hand side, and let $D_2$ be the triangle that you obtain by reflecting $D_1$ across the line $y=x$. Then $\iint_{D_1} f(x) \, f(y) \, dx dy = \iint_{D_2} f(x) \, f(y) \, dx dy$ since the integrand $g(x,y) = f(x) \, f(y)$ is symmetric with respect to the reflection: $g(y,x)=g(x,y)$. Together, the triangles $D_1$ and $D_2$ make up the square $E = [a,b]^2$, and they overlap only along the diagonal line, which has measure zero, so $\iint_{D_1} g + \iint_{D_2} g = \iint_E g$.
