The area of the region given by {(x, y): xy l | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The region is bounded by $y = 1$,$y = x^2$,and $xy = 8$ (or $y = 8/x$).First,find the intersection points:$x^2 = 1 \implies x = 1$ (for $x > 0$).$x^2 = 8/x \implies x^3 = 8 \implies x = 2$.$8/x = 1 \implies x = 8$.The area is given by the sum of two integrals:Area $= \int \limits_1^2 (x^2 - 1) dx + \int \limits_2^8 (8/x - 1) dx$$= \left[ \frac{x^3}{3} - x \right]_1^2 + \left[ 8 \ln|x| - x \right]_2^8$$= \left( (8/3 - 2) - (1/3 - 1) \right) + \left( (8 \ln 8 - 8) - (8 \ln 2 - 2) \right)$$= (2/3 - (-2/3)) + (8(3 \ln 2) - 8 - 8 \ln 2 + 2)$$= 4/3 + 24 \ln 2 - 8 \ln 2 - 6$$= 16 \ln 2 + 4/3 - 6$$= 16 \ln 2 - 14/3$.

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

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