For $X_1, X_2 \sim Unif(0,1)$ independent, why is $P(X_1X_2 >k) = \int_{k}^{1}\int_{k/x_1}^{1}dx_2dx_1$?

Question

For $X_1, X_2 \sim Unif(0,1)$ independent, I am trying to see why it is the case that

$$ P(X_1X_2 >k) = \int_{k}^{1}\int_{k/x_1}^{1}dx_2dx_1 $$

I get the inner integral bound, but I do not know why the second outer integral starts from $k$ instead of starting from $0$?

Answer 1

The largest $x_2$ can be is $1$, so $1 \geqslant x_2 $, or $1/x_2 \geqslant 1$. Therfore in order to have $x_1x_2>k$, you must have $$ x_1 > \frac{k}{x_2} \geqslant k. $$

Quick sketch of the region in question: