Question

Evaluate : ${\int }_0^{3}f\left(x\right)dx$, where $f\left(x\right)=\{\begin{array}{c}\cos 2x0\le x\le \frac{\pi }{2}\\ 3\frac{\pi }{2}\le x\le 3\end{array}$

Answer 1

${\int }_0^{3}f\left(x\right)dx$

$={\int }_0^{\frac{\pi }{2}}f\left(x\right)dx+{\int }_{\frac{\pi }{2}}^{3}f\left(x\right)dx$

$={\int }_0^{\frac{\pi }{2}}\cos 2xdx+{\int }_{\frac{\pi }{2}}^{3}3dx$

$={\left[\begin{array}{c}\frac{\sin 2x}{2}\end{array}\right]}_0^{\pi /2}+[3x{]}_{\frac{\pi }{2}}^3$

$=\left[\begin{array}{c}\frac{\sin 2\times \pi /2}{2}-\frac{\sin 2\times 0}{2}\end{array}\right]+[3\times 3-\frac{3\pi }{2}]$

$=\frac{\sin \pi }{2}-0+9-\frac{3\pi }{2}$ $(\because \sin \pi =0)$

$=9-\frac{3\pi }{2}=3\left(\begin{array}{c}3-\frac{\pi }{2}\end{array}\right)$