## Calculus 8th Edition

Note that $\int_{0}^{1} f(x) \, dx = \int_{0}^{1} f(y) \, dy$ as the area of integration of $f$ stays the same with both equations. Thus, $\int_{0}^{1} \int_{0}^{1} f(x) f(y) \, dy \, dx = \int_{0}^{1} f(x) \, dx \int_{0}^{1} f(y) \, dy = \left[ \int_{0}^{1} f(x) \, dx \right]^2$.