The area of the region bounded by the curve y | vedclass

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Question

(B) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.Since the region is bounded by the $Y$-axis $(x = 0)$,the curve $x = \frac{y^2}{4}

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(B) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.Since the region is bounded by the $Y$-axis $(x = 0)$,the curve $x = \frac{y^2}{4}$,and the line $y = 3$,the area $A$ is given by the integral with respect to $y$ from $y = 0$ to $y = 3$.$A = \int_{0}^{3} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$.$A = \frac{1}{4} \left[ \frac{y^3}{3} ight]_{0}^{3}$.$A = \frac{1}{4} \left( \frac{3^3}{3} - 0 ight) = \frac{1}{4} 	imes \frac{27}{3} = \frac{9}{4}$ sq. units.Therefore,the correct option is $B$.

The area of the region bounded by the curve $y^2 = 4x$,the $Y$-axis,and the line $y = 3$ is . . . . . . sq. units.

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