Question

Find the area of the region enclosed by the curves $y=\sin \pi x,y=x^2-x,x=2$.

Answer 1

$\begin{array}{c}Requiredarea={\int }_1^2{x}^2+xdx+{\int }_1^2\sin xdx\\ ={\left[\frac{x^3}{3}+\frac{x^2}{2}\right]}_1^2+{\left[\frac{\cos \pi x}{\pi }\right]}_1^2\\ =\frac{8}{3}+2-\left(\frac{1}{3}+\frac{1}{2}\right)+\left(\frac{1}{\pi }-\frac{1}{\pi }\right)\\ =\frac{14}{3}-\frac{5}{6}=\frac{23}{6}sq.unit\end{array}$