find the area enclosed by y^2=2x+1 & x-y-1=0

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{x}=\mathrm{y}+1 ;\left(\mathrm{i}\right) \\ {\mathrm{y}}^2=2\mathrm{x}+1 \\ \mathrm{By} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \Rightarrow {\left(\mathrm{x}-1\right)}^2=2\mathrm{x}+1 \\ \Rightarrow {\mathrm{x}}^2+1-2\mathrm{x}=2\mathrm{x}+1 \\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}-4\right)=0 \\ \Rightarrow \mathrm{x}=0 \mathrm{and} \mathrm{x}=4 \\ \mathrm{Put} \mathrm{x}=0 \mathrm{in} \left(\mathrm{i}\right) \\ 0=\mathrm{y}+1 \\ \Rightarrow \mathrm{y}=-1 \\ \mathrm{Put} \mathrm{x}=4 \mathrm{in} \left(\mathrm{i}\right) \\ 4=\mathrm{y}+1 \\ \Rightarrow \mathrm{y}=3 \\ \left(\mathrm{ii}\right) \mathrm{can} \mathrm{be} \mathrm{rewritten} \mathrm{as} : \\ {\mathrm{y}}^2=2\mathrm{x}+1 \\ \Rightarrow \mathrm{x}=\frac{1}{2}\left({\mathrm{y}}^2-1\right) \\ \mathrm{Hence} \mathrm{required} \mathrm{area} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{A}=\left|{\int }_{-1}^3\left[\left(\mathrm{y}+1\right)-\frac{1}{2}\left({\mathrm{y}}^2-1\right)\right]\mathrm{dy}\right| \\ =\left|{\int }_{-1}^3\left[\mathrm{y}+1-\frac{1}{2}{\mathrm{y}}^2+\frac{1}{2}\right]\mathrm{dy}\right| \\ =\left|{\int }_{-1}^3\left[\mathrm{y}-\frac{{\mathrm{y}}^2}{2}+\frac{3}{2}\right]\mathrm{dy}\right| \\ =\left|{\left[\frac{{\mathrm{y}}^2}{2}-\frac{{\mathrm{y}}^3}{6}+\frac{3}{2}\mathrm{y}\right]}_{-1}^3\right| \\ =\left|\left[\frac{9}{2}-\frac{27}{6}+\frac{9}{2}\right]-\left[\frac{1}{2}+\frac{1}{6}-\frac{3}{2}\right]\right| \\ =\left|\left[9-\frac{9}{2}\right]-\left[\frac{1}{6}-1\right]\right| \\ =\left|\frac{9}{2}-\frac{1}{6}+1\right| \\ =\left|\frac{27-1+6}{6}\right| \\ =\left|\frac{32}{6}\right| \\ =\frac{16}{3} {\mathrm{unit}}^2 \end{gathered}$$
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