The area (in sq. units) bounded by the curves | vedclass

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(A) First,we find the intersection point of the curves $y = \frac{8}{x}$ and $y = 2x$.Setting $\frac{8}{x} = 2x$,we get $x^2 = 4$,so $x = 2$ (since $x

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$12-8 \log 2$

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(A) First,we find the intersection point of the curves $y = \frac{8}{x}$ and $y = 2x$.Setting $\frac{8}{x} = 2x$,we get $x^2 = 4$,so $x = 2$ (since $x > 0$ in the first quadrant).The region is bounded by $x = 2$ to $x = 4$.In this interval,$2x \geq \frac{8}{x}$.The required area is given by the integral:$	ext{Area} = \int_2^4 \left( 2x - \frac{8}{x} ight) dx$$= \left[ x^2 - 8 \log |x| ight]_2^4$$= (4^2 - 8 \log 4) - (2^2 - 8 \log 2)$$= (16 - 8 \log 4) - (4 - 8 \log 2)$$= 12 - 8 \log 4 + 8 \log 2$$= 12 - 8 \log (2^2) + 8 \log 2$$= 12 - 16 \log 2 + 8 \log 2$$= 12 - 8 \log 2$

The area (in sq. units) bounded by the curves $y=\frac{8}{x}$,$y=2x$ and $x=4$ is

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