Mathematics - Area Under Curves Question with Solution | TestHub

We have, $y=x+1\text{ and }y=\cos x$$\begin{array}{rl}\text{ Require area }& ={\int }_{-1}^0(x+1)dx+{\int }_0^{\pi /2}\cos xdx\\ & ={\left[\frac{x^2}{2}+x\right]}_{-1}^0+[\sin x{]}_0^{\pi /2}\\ & =\left[(0)-\left(\frac{1}{2}-1\right)\right]+[1-0]\\ & =\frac{1}{2}+1=\frac{3}{2}\text{ sq unit. }\end{array}$