Area of the region bounded by x^2 = 4y,the X- | vedclass

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(B) The given equation of the parabola is $x^2 = 4y$,which implies $y = \frac{x^2}{4}$.To find the area bounded by the curve $y = \frac{x^2}{4}$,the $

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$\frac{9}{4}$

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(B) The given equation of the parabola is $x^2 = 4y$,which implies $y = \frac{x^2}{4}$.To find the area bounded by the curve $y = \frac{x^2}{4}$,the $X$-axis,and the line $x = 3$,we integrate the function with respect to $x$ from $x = 0$ to $x = 3$.Area $= \int_{0}^{3} y \, dx = \int_{0}^{3} \frac{x^2}{4} \, dx$.Area $= \frac{1}{4} \left[ \frac{x^3}{3} ight]_{0}^{3}$.Area $= \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} ight) = \frac{1}{4} \left( \frac{27}{3} ight) = \frac{1}{4} 	imes 9 = \frac{9}{4}$ sq. units.Therefore,the correct option is $B$.

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

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