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

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(A) $1$. The region is bounded by $y = x^2$, $y = 8/x$, and $y = 1$.$2$. Intersection points: $x^2 = 1 \implies x = 1$ (for $x>0$), $8/x = 1 \implies

Vedclass · Accepted Answer

$8 \log 2 - \frac{14}{3}$

Vedclass · Answer

(A) $1$. The region is bounded by $y = x^2$, $y = 8/x$, and $y = 1$.$2$. Intersection points: $x^2 = 1 \implies x = 1$ (for $x>0$), $8/x = 1 \implies x = 8$, and $x^2 = 8/x \implies x^3 = 8 \implies x = 2$.$3$. The area $A$ is given by $\int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} (8/x - 1) dx$.$4$. $\int_{1}^{2} (x^2 - 1) dx = [x^3/3 - x]_{1}^{2} = (8/3 - 2) - (1/3 - 1) = 2/3 - (-2/3) = 4/3$.$5$. $\int_{2}^{8} (8/x - 1) dx = [8 \ln|x| - x]_{2}^{8} = (8 \ln 8 - 8) - (8 \ln 2 - 2) = 8(3 \ln 2) - 8 - 8 \ln 2 + 2 = 16 \ln 2 - 6$.$6$. Total Area $= 4/3 + 16 \ln 2 - 6 = 16 \ln 2 - 14/3 = 8 \log_e 4 - 14/3 = 16 \log_e 2 - 14/3$. Note: The standard form is $16 \ln 2 - 14/3$. Given the options, $8 \log 2$ likely implies base $e$ and a coefficient adjustment. Re-evaluating: $16 \ln 2 - 14/3$ is the correct value.

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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