Area of the region bounded by $\{(x,y):0\le y\le x^2+1,0\le y\le x+1,0\le x\le 2\}$  
 

$$\begin{gathered} \begin{array}{l}y=x^2+1\&y=x+1\\ x^2+1=x+1\\ x^2-x=0\\ x=o,x=1\end{array} \\ The point of intersection is (0,1)\&(1,2) \\ \begin{array}{l}y=x+1\\ putx=2\\ y=3\therefore (2,3),x=2\\ Area\underset{0}{\overset{1}{\int }}(x^2+1)dx+\underset{1}{\overset{2}{\int }}(x+1)dx\\ =\frac{x^3}{3}+x{{|{{ }_0}^1+\frac{x^2}{2}+x|}_1}^2\end{array} \\ \begin{array}{l}=(\frac{1}{3}+1)+[(\frac{4}{2}+2)-(\frac{1}{2}+1)]\\ \Rightarrow \frac{4}{3}+[4-\frac{3}{2}]=\frac{4}{3}+\frac{5}{2}=\frac{8+15}{6}=\frac{23}{6}uni{t}^2\end{array} \end{gathered}$$