The area (in sq. units) of the region enclose | vedclass

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(D) The region is bounded by $y = x^2$, $y = 8/x$, and $y = 1$.Find intersection points:$1$. $y = x^2$ and $y = 1 \implies x = \pm 1$. Since $xy \leq

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$16 \log 2 - \frac{14}{3}$

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(D) The region is bounded by $y = x^2$, $y = 8/x$, and $y = 1$.Find intersection points:$1$. $y = x^2$ and $y = 1 \implies x = \pm 1$. Since $xy \leq 8$ and $y \geq 1$, we consider $x > 0$, so $x = 1$.$2$. $y = 8/x$ and $y = 1 \implies x = 8$.$3$. $y = x^2$ and $y = 8/x \implies x^3 = 8 \implies x = 2$.The area $A = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} (8/x - 1) dx$.$A = [x^3/3 - x]_{1}^{2} + [8 \ln|x| - x]_{2}^{8}$.$A = (8/3 - 2) - (1/3 - 1) + (8 \ln 8 - 8) - (8 \ln 2 - 2)$.$A = (2/3) - (-2/3) + 8(3 \ln 2) - 8 - 8 \ln 2 + 2$.$A = 4/3 + 24 \ln 2 - 8 - 8 \ln 2 + 2 = 16 \ln 2 - 14/3$.

The area (in sq. units) of the region enclosed by the set of points $\{(x, y) : y \leq x^2, xy \leq 8, y \geq 1\}$ is:

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