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(B) Given the curve $y = 4x^2$,we can express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.Since the region is in the first qu

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$\frac{1}{3}[8-2\sqrt{2}]$ sq. units

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(B) Given the curve $y = 4x^2$,we can express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.Since the region is in the first quadrant and bounded by $x=0$,$y=2$,and $y=4$,the area $A$ is given by the integral with respect to $y$:$A = \int_{2}^{4} x \, dy = \int_{2}^{4} \frac{\sqrt{y}}{2} \, dy$.Evaluating the integral:$A = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} \left[ y^{3/2} \right]_{2}^{4}$.$A = \frac{1}{3} [4^{3/2} - 2^{3/2}] = \frac{1}{3} [8 - 2\sqrt{2}]$ sq. units.

The area of the region lying in the first quadrant bounded by $y=4x^2$,$x=0$,$y=2$,and $y=4$ is

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