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

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(C) The given curve is $y^2 = 4x$ and the line is $x = 3$.Since the curve is symmetric about the $x$-axis,the total area $A$ is given by $2 	imes \in

Vedclass · Accepted Answer

$8 \sqrt{3}$

Vedclass · Answer

(C) The given curve is $y^2 = 4x$ and the line is $x = 3$.Since the curve is symmetric about the $x$-axis,the total area $A$ is given by $2 	imes \int_{0}^{3} y \, dx$.From $y^2 = 4x$,we have $y = \sqrt{4x} = 2\sqrt{x}$.Thus,$A = 2 \int_{0}^{3} 2\sqrt{x} \, dx = 4 \int_{0}^{3} x^{1/2} \, dx$.Evaluating the integral: $A = 4 \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{3} = 4 	imes \frac{2}{3} 	imes [x^{3/2}]_{0}^{3}$.$A = \frac{8}{3} 	imes (3)^{3/2} = \frac{8}{3} 	imes 3\sqrt{3} = 8\sqrt{3}$ sq. units.Therefore,the correct option is $C$.

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

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