Mathematics - Area Under Curves Question with Solution | TestHub

$\begin{array}{rl}& \text{ Required area }={\int }_{-1}^3\left\{(1+y)-\left(\frac{y^2-1}{2}\right)\right\}\mathrm{d}y\\ & ={\int }_{-1}^3\left(\frac{3}{2}+y-\frac{y^2}{2}\right)\mathrm{d}y\\ & ={\left[\frac{3}{2}y+\frac{y^2}{2}-\frac{y^3}{6}\right]}_{-1}^3\\ & =\left(\frac{9}{2}+\frac{9}{2}-\frac{27}{6}\right)-\left(\frac{-3}{2}+\frac{1}{2}+\frac{1}{6}\right)=\frac{16}{3}\end{array}$