Find area of region { (x,y) : 0<y< x²+1, 0< y < x+1, 0<x<2 }

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \left\{\left(\mathrm{x},\mathrm{y}\right): 0<\mathrm{x}<{\mathrm{y}}^2+1, 0<\mathrm{y}<\mathrm{x}+1, 0<\mathrm{x}<2\right\} \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{common} \mathrm{area} \mathrm{is} \mathrm{represented} \mathrm{in} \mathrm{the} \mathrm{graph}. \\ \mathrm{Now} \\ \mathrm{Finding} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{is} \mathrm{easy}. \\ \mathrm{Put} \\ {\mathrm{x}}^2+1=\mathrm{x}+1 \\ \Rightarrow {\mathrm{x}}^2=\mathrm{x} \\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}-1\right)=0 \\ \Rightarrow \mathrm{x}=0 \mathrm{and} \mathrm{x}=1 \\ \mathrm{which} \mathrm{are} \mathrm{point} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{in} \mathrm{graph}. \\ \mathrm{For} \mathrm{C} , \mathrm{we} \mathrm{have} \mathrm{x}=2 \\ \mathrm{Hence} \mathrm{required} \mathrm{area} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}: \\ \mathrm{Area}={\int }_0^1\left({\mathrm{x}}^2+1\right).\mathrm{dx}+{\int }_1^2\left(\mathrm{x}+1\right)\mathrm{dx} \\ ={\left[\frac{{\mathrm{x}}^3}{3}+\mathrm{x}\right]}_0^1+{\left[\frac{{\left(\mathrm{x}+1\right)}^2}{2}\right]}_1^2 \\ =\left[\left(\frac{1}{3}+1\right)-0\right]+\left[\frac{9}{2}-\frac{4}{2}\right] \\ =\frac{4}{3}+\frac{5}{2} \\ =\frac{8+15}{6} \\ =\frac{\mathbf{23}}{\mathbf{6}}\mathbf{ }{\mathbf{unit}}^{\mathbf{2}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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