The area of the region given by (x,y):xy≤8, 1≤y≤x2 is:

Given curvesC1:y=8xC2:y=x2C3:y=1Point of intersection of C1 and C2 is (2, 4) C3 and C2 is (1, 1) C1 and C3 is (8, 1)Required areas =∫12 x2−1dx+∫28 8x−1dx=x33−x12+8lnx−x28=73−1+8ln4−6=16⋅ln2−143

The area of the region given by (x,y):xy≤8, 1≤y≤x2 is:

Given curvesC1:y=8xC2:y=x2C3:y=1Point of intersection of C1 and C2 is (2, 4) C3 and C2 is (1, 1) C1 and C3 is (8, 1)Required areas =∫12 x2−1dx+∫28 8x−1dx=x33−x12+8lnx−x28=73−1+8ln4−6=16⋅ln2−143

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The area of the region given by $\{(x, y) : xy \leq 8, 1 \leq y \leq x^{2}\}$ is:

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$16 \log_{e} 2 + \frac{7}{3}$

$16 \log_{e} 2 - \frac{14}{3}$

$8 \log_{e} 2 - \frac{13}{3}$

$8 \log_{e} 2 + \frac{7}{6}$

answer is A.

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Detailed Solution

Given curves
$\begin{matrix} C_{1} : y = \frac{8}{x} \\ C_{2} : y = x^{2} \\ C_{3} : y = 1 \end{matrix}$
Point of intersection of C₁ and C₂ is (2, 4)
C₃ and C₂ is (1, 1)
C₁ and C₃ is (8, 1)
Required areas $= \int_{1}^{2} (x^{2} - 1) dx + \int_{2}^{8} (\frac{8}{x} - 1) dx$
$\begin{array}{l} = {(\frac{x^{3}}{3} - x)}_{1}^{2} + {(8 lnx - x)}_{2}^{8} \\ = (\frac{7}{3} - 1) + 8 \ln 4 - 6 \\ = 16 \cdot \ln 2 - \frac{14}{3} \end{array}$