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(A) Given curve is $y = 4x^2$, which implies $x^2 = \frac{y}{4}$, so $x = \frac{\sqrt{y}}{2}$ (since it is in the first quadrant).Area $A = \int_{2}^{

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

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(A) Given curve is $y = 4x^2$, which implies $x^2 = \frac{y}{4}$, so $x = \frac{\sqrt{y}}{2}$ (since it is in the first quadrant).Area $A = \int_{2}^{4} x \, dy$.$A = \int_{2}^{4} \frac{\sqrt{y}}{2} \, dy$.$A = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy$.$A = \frac{1}{2} [\frac{y^{3/2}}{3/2}]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} [y^{3/2}]_{2}^{4}$.$A = \frac{1}{3} [4^{3/2} - 2^{3/2}]$.$A = \frac{1}{3} [8 - 2\sqrt{2}]$ sq. units.

The area of the region lying in the first quadrant and bounded by the curve $y = 4x^2$, the $Y$-axis, and the lines $y = 2$ and $y = 4$ is:

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