Proof for an integral identity Is it true that $\int_0^A dx \int_0^B dy f(x) f(y) = 2 \int_0^A dx \int_0^x dy f(x) f(y)$ ? If so, can this be proved?
 A: Let's assume for simplicity that $f$ has an anti-derivative $F$. To simplify notation, write $F_t = F(t)$. Then the left-hand side yields
$$\int_0^A f(x) dx \times \int_0^B f(y) dy = (F_A-F_0)\cdot (F_B-F_0)$$
while the right-hand side becomes
$$\begin{split}
2 \int_0^A f(x)dx \int_0^x f(y) dy
 &= 2 \int_0^A f(x) (F_x - F_0)dx \\
 &= 2 \int_0^A f(x) F_x dx - 2F_0 (F_A - F_0) \\
 &= (F_A^2 - F_0^2) - 2F_0 (F_A - F_0) \\
 &= (F_A - F_0) \cdot (F_A + F_0 - 2F_0) \\
 &= (F_A - F_0)^2
\end{split}
$$
So these coincide if and only if $F_A = F_B$ or $F_A=F_0$.
EDIT A clarification comment. $\int F_x f(x) dx$ is integrated by subtitution with $u=F_x$ and then $du = f(x) dx$.
A: Yes, it's true by symmetry. The function $F(x,y)=f(x)f(y)$ is symmetric about the line $y=x$. The first integral is over the square, the second is over the bottom triangle. To give a formula, add $\int_0^1 dy\int_0^y dx$, which is clearly equal to the given integral on the RHS.
A: My answer is essentially the same as gt6989b's, but I will leave it in case the difference in presentation is useful.
Define
$$
F(x)=\int_0^xf(y)\,\mathrm{d}y
$$
Then $f(x)=F'(x)$ and
$$
\begin{align}
2\int_0^A\int_0^x f(x)f(y)\,\mathrm{d}y\,\mathrm{d}x
&=2\int_0^Af(x)F(x)\,\mathrm{d}x\\
&=2\int_0^AF'(x)F(x)\,\mathrm{d}x\\
&=F(A)^2\\
&=\left(\int_0^Af(x)\,\mathrm{d}x\right)^2
\end{align}
$$
However,
$$
\int_0^A\int_0^B f(x)f(y)\,\mathrm{d}y\,\mathrm{d}x
=\left(\int_0^Af(x)\,\mathrm{d}x\right)\left(\int_0^Bf(y)\,\mathrm{d}y\right)
$$
Thus, your equation is true if
$$
\int_0^Af(x)\,\mathrm{d}x=\int_0^Bf(y)\,\mathrm{d}y
$$
or if
$$
\int_0^Af(x)\,\mathrm{d}x=0
$$
