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(A) The region bounded by the parabola $y^{2}=4x$ and the line $x=3$ is the area $OAC$ (as shown in the figure).The area is symmetrical about the $x$-

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$8$

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(A) The region bounded by the parabola $y^{2}=4x$ and the line $x=3$ is the area $OAC$ (as shown in the figure).The area is symmetrical about the $x$-axis.Therefore,the required area $= 2 	imes (	ext{Area of } OAB)$.Area $= 2 \int_{0}^{3} y \, dx$Since $y^{2}=4x$,we have $y = 2\sqrt{x}$ (taking the positive value for the upper half).Area $= 2 \int_{0}^{3} 2\sqrt{x} \, dx = 4 \int_{0}^{3} x^{1/2} \, dx$$= 4 \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{3} = 4 	imes \frac{2}{3} \left[ x^{3/2} ight]_{0}^{3}$$= \frac{8}{3} \left[ 3^{3/2} - 0^{3/2} ight] = \frac{8}{3} 	imes 3\sqrt{3} = 8\sqrt{3}$ square units.

Find the area of the region bounded by the curve $y^{2}=4x$ and the line $x=3$. (in $\sqrt{3}$)

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