Area of the region bounded by the curve y^2 = | vedclass

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Question

(C) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.We need to find the area bounded by this curve,the $Y$-axis $(x = 0)$,and the line

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

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

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

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