The area (in square units) bounded by the cur | vedclass

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(A) The region is bounded by the square defined by $x = \pm 2$ and $y = \pm 2$, which has a total area of $4 	imes 4 = 16$. The condition $xy \leq \f

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$9+3 \log 2$

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(A) The region is bounded by the square defined by $x = \pm 2$ and $y = \pm 2$, which has a total area of $4 	imes 4 = 16$. The condition $xy \leq \frac{1}{2}$ excludes the region where $xy > \frac{1}{2}$.This region $xy > \frac{1}{2}$ consists of two parts: one in the first quadrant where $y > \frac{1}{2x}$ and one in the third quadrant where $y Due to symmetry, the area of these two parts is equal. Let's calculate the area in the first quadrant bounded by $x=2, y=2, x=1/4$ (since $2x=1/2 \implies x=1/4$ at $y=2$) and the curve $y=1/(2x)$.The area in the first quadrant where $xy > 1/2$ is $\int_{1/4}^{2} (2 - \frac{1}{2x}) dx = [2x - \frac{1}{2} \log x]_{1/4}^{2} = (4 - \frac{1}{2} \log 2) - (1/2 - \frac{1}{2} \log(1/4)) = 4 - 0.5 \log 2 - 0.5 + 0.5 \log(2^{-2}) = 3.5 - 0.5 \log 2 - \log 2 = 3.5 - 1.5 \log 2$.Total area to exclude = $2 	imes (3.5 - 1.5 \log 2) = 7 - 3 \log 2$.Required area = Total area of square - Excluded area = $16 - (7 - 3 \log 2) = 9 + 3 \log 2$.

The area (in square units) bounded by the curves $|x|=2, |y|=2$ and $xy \leq \frac{1}{2}$ is

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