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(A) The area in the first quadrant bounded by $y=4x^2$,$x=0$,$y=1$,and $y=4$ is calculated by integrating with respect to $y$.Given $y=4x^2$,we have $

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$\frac{7}{3}$ sq. units

Vedclass · Answer

(A) The area in the first quadrant bounded by $y=4x^2$,$x=0$,$y=1$,and $y=4$ is calculated by integrating with respect to $y$.Given $y=4x^2$,we have $x^2 = \frac{y}{4}$,which implies $x = \frac{\sqrt{y}}{2}$ (since $x \ge 0$ in the first quadrant).The area $A$ is given by the integral:$A = \int_{1}^{4} x \, dy$$A = \int_{1}^{4} \frac{\sqrt{y}}{2} \, dy$$A = \frac{1}{2} \int_{1}^{4} y^{1/2} \, dy$$A = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} ight]_{1}^{4}$$A = \frac{1}{2} \cdot \frac{2}{3} \left[ y^{3/2} ight]_{1}^{4}$$A = \frac{1}{3} [4^{3/2} - 1^{3/2}]$$A = \frac{1}{3} [8 - 1]$$A = \frac{7}{3} 	ext{ sq. units}$.

Find the area of the region lying in the first quadrant and bounded by $y=4x^2$,$x=0$,$y=1$,and $y=4$.

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