The area bounded by the curve $y^2=2 x+1$ and the line $x-y=1$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Required area }=\int_{-1}^3\left\{(1+y)-\left(\frac{y^2-1}{2}ight)ight\} \mathrm{d} y \ & =\int_{-1}^3\left(\frac{3}{2}+y-\frac{y^2}{2}ight) \mathrm{d} y \ & =\left[\frac{3}{2} y+\frac{y^2}{2}-\frac{y^3}{6}ight]_{-1}^3 \ & =\left(\frac{9}{2}+\frac{9}{2}-\frac{27}{6}ight)-\left(\frac{-3}{2}+\frac{1}{2}+\frac{1}{6}ight)=\frac{16}{3}\end{aligned}$

The area bounded by the curve $y^2=2 x+1$ and the line $x-y=1$ is

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