Proving a double integral

Question

How does $2\int_a^b\int_x^b f(x) f(y) dydx = (\int_a^b f(x)dx)^2$?

I know that $(\int_a^b f(x)dx)^2 = \int\int_{[a,b] \times [a,b]} f(x) f(y) dxdy$ but I don't get how I can get to $2\int_a^b\int_x^b f(x) f(y) dydx$ from that conclusion.

Answer 1

The limits are $a \leq x \leq b$, $x \leq y \leq b$, which is equivalent to $a \leq y \leq b$, $a \leq x \leq y$. So, $$\begin{align}2\int_a^b\int_x^b f(x) f(y) \,dy\,dx &= \int_a^b\int_x^b f(x) f(y) \,dy\,dx+\int_a^b\int_a^y f(x) f(y) \,dx\,dy \\ &=\int_a^b\int_x^b f(x) f(y) \,dy\,dx+\int_a^b\int_a^x f(y) f(x) \,dy\,dx \\ &=\int_a^b\left(\int_x^b f(x) f(y) \,dy+\int_a^x f(y) f(x) \,dy\right) \,dx\\ &=\int_a^b f(x)\left(\int_x^b f(y) \,dy+\int_a^x f(y) \,dy\right) \,dx\\ &=\int_a^b f(x) \left(\int_x^b f(y) \,dy+\int_a^x f(y) \,dy\right) \,dx\\ &=\int_a^b f(x) \left(\int_a^b f(y) \,dy\right) \,dx\\ &=\left(\int_a^b f(x) \, dx \right)\left(\int_a^b f(y) \, dy \right)\\ &= \left(\int_a^b f(x)\right)^2\end{align}$$

Answer 2

If you draw a picture of the domain of integration on the left-hand side, you'll see it. It's just the integral over half a rectangle, and because of the nature of the integrand, that's half the integral over the rectangle.