The area of the region bounded by the curve $y^2=9 x$ and the line $y=3 x$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Required area }=\int_0^1(3 \sqrt{x}-3 x) d x \ & =3\left[\frac{2}{3} x^{3 / 2}-\frac{x^2}{2}ight]_0^1 \ & =3\left(\frac{2}{3}-\frac{1}{2}ight) \ & =3 	imes \frac{1}{6}=\frac{1}{2}\end{aligned}$

The area of the region bounded by the curve $y^2=9 x$ and the line $y=3 x$ is

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