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

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Aicharya · Accepted Answer

Refer figure, point of intersection of given curves are $\mathrm{x}^2=4 \mathrm{x}$ $\Rightarrow \mathrm{x}(\mathrm{x}-4)=0$ $	herefore \mathrm{O} \equiv(0,0) 	ext { and } \mathrm{P} \equiv(4,4)$ Required area is shaded. $\begin{aligned} & 	herefore A=\int_0^4(\sqrt{4 x}-x) d x=2 \int_0^4 x^{\frac{1}{2}} d x-\int_0^4 x d x \ & =2\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}ight)}ight]_0^4-\left[\frac{x^2}{2}ight]_0^4-\left(\frac{4}{3}ight)(4 	imes 2)-\frac{16}{2} \ & =\frac{32}{3}-\frac{16}{2}=\frac{16}{6}=\frac{8}{3} 	ext { sq. units. } \end{aligned}$

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

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