The area of the region R={(x,y):xy le 8, 1 le | vedclass

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(C) The region $R$ is bounded by $y=x^2$, $y=1$, and $xy=8$ (or $y=8/x$).First, find the intersection points:$x^2=1 \implies x=1$ (since $x \ge 0$)$x^

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$\frac{2}{3}(24\,\log_{e}(2)-7)$

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(C) The region $R$ is bounded by $y=x^2$, $y=1$, and $xy=8$ (or $y=8/x$).First, find the intersection points:$x^2=1 \implies x=1$ (since $x \ge 0$)$x^2=8/x \implies x^3=8 \implies x=2$$8/x=1 \implies x=8$The area $A$ is given by the sum of two integrals:$A = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} (\frac{8}{x} - 1) dx$Evaluating the first integral:$\int_{1}^{2} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{1}^{2} = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$Evaluating the second integral:$\int_{2}^{8} (\frac{8}{x} - 1) dx = [8 \ln|x| - x]_{2}^{8} = (8 \ln 8 - 8) - (8 \ln 2 - 2) = 8(3 \ln 2) - 8 - 8 \ln 2 + 2 = 24 \ln 2 - 8 \ln 2 - 6 = 16 \ln 2 - 6$Total area $A = \frac{4}{3} + 16 \ln 2 - 6 = 16 \ln 2 - \frac{14}{3} = \frac{48 \ln 2 - 14}{3} = \frac{2}{3}(24 \ln 2 - 7)$.

The area of the region $R=\{(x,y):xy \le 8, 1 \le y \le x^{2}, x \ge 0\}$ is

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