Answer
$P\left( {XY \ge \frac{1}{2}} \right) \simeq 0.153$
Work Step by Step
We have the joint probability density function:
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
1&{{\rm{if}}{\ }0 \le x \le 1,0 \le y \le 1}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
Notice that the domain ${\cal D}$ of the $p\left( {x,y} \right)$ for nonzero $p$ is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 1} \right\}$
To calculate the probability that $XY \ge \frac{1}{2}$, we must define the domain ${{\cal D}_1}$ such that $xy \ge \frac{1}{2}$ and satisfies the domain of $p\left( {x,y} \right)$ where $p$ is nonzero, that is, $0 \le x \le 1$, $0 \le y \le 1$.
We sketch the domain and see that we can describe ${{\cal D}_1}$ as a vertically simple region, bounded below by the curve $xy = \frac{1}{2}$ and bounded above by the line $y=1$.
We find the left boundary by intersecting the curve $xy = \frac{1}{2}$ and the line $y=1$:
$x\cdot 1 = \frac{1}{2}$, ${\ \ \ \ }$ $x = \frac{1}{2}$
So, the range of $x$ is $\frac{1}{2} \le x \le 1$.
Thus, the description of ${{\cal D}_1}$:
${{\cal D}_1} = \left\{ {\left( {x,y} \right)|\frac{1}{2} \le x \le 1,\frac{1}{{2x}} \le y \le 1} \right\}$
Evaluate:
$P\left( {XY \ge \frac{1}{2}} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 1/2}^1 \mathop \smallint \limits_{y = 1/\left( {2x} \right)}^1 {\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 1/2}^1 \left( {y|_{1/\left( {2x} \right)}^1} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 1/2}^1 \left( {1 - \frac{1}{{2x}}} \right){\rm{d}}x$
$ = \left( {x - \frac{1}{2}\ln x} \right)|_{1/2}^1$
$ = 1 - \frac{1}{2} + \frac{1}{2}\ln \frac{1}{2} = \frac{1}{2} - \frac{1}{2}\ln 2 \simeq 0.153$
So, $P\left( {XY \ge \frac{1}{2}} \right) \simeq 0.153$.
