The area of the region lying in the first qua | vedclass

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Question

(A) Given the curve $y = 4x^2$, we can express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.The area $A$ in the first quadrant

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the curve $y = 4x^2$, we can express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.The area $A$ in the first quadrant bounded by the $Y$-axis and the lines $y = 2$ and $y = 4$ is given by the integral:$A = \int_{2}^{4} x \, dy = \int_{2}^{4} \frac{\sqrt{y}}{2} \, dy$.$A = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy$.$A = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} ight]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} \left[ y^{3/2} ight]_{2}^{4}$.$A = \frac{1}{3} [4^{3/2} - 2^{3/2}] = \frac{1}{3} [8 - 2\sqrt{2}] 	ext{ sq. units}$.

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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