Mathematics - Definite Integration Question with Solution | TestHub

Given, ${\int }_{-3}^{101}\left(\left[\sin \pi x\right]+e^{\left[\cos 2\pi x\right]}\right)dx$

Now let $I_1={\int }_{-3}^{101}\left[\sin \pi x\right]dx$ and $I_2={\int }_{-3}^{101}{e}^{\left[\cos 2\pi x\right]}dx$

Now checking periodicity of $\left[\sin \pi x\right]$ we get,

$0<x<1\Rightarrow \left[\sin \pi x\right]=0$

$1<x<2\Rightarrow \left[\sin \pi x\right]=-1$

$\left[\sin \pi x\right]\to$ periodic with period 2

So, $I_1=52{\int }_0^2\left[\sin \pi x\right]dx$

$=52\left[{\int }_0^{1}0+{\int }_1^2-1dx\right]=-52$

Now checking periodicity of $\left[\cos 2\pi x\right]$ we get,

$0<x<\frac{1}{4}\Rightarrow \left[\cos 2\pi x\right]=0$

$\frac{1}{4}<x<\frac{1}{2}\Rightarrow \left[\cos 2\pi x\right]=-1$

$\frac{1}{2}<x<\frac{3}{4}\Rightarrow \left[\cos 2\pi x\right]=-1$

$\frac{3}{4}<x<1\Rightarrow \left[\cos 2\pi x\right]=0$

So, $\left[\cos 2\pi x\right]$ has periodicity of $1$

So, $I_2=104\left({\int }_0^{\frac{1}{4}}{e}^{0}dx+{\int }_{\frac{1}{4}}^{\frac{3}{4}}{e}^{-1}dx+{\int }_{\frac{3}{4}}^1{e}^{0}dx\right)$

$\Rightarrow I_2=104\left(\frac{1}{4}+\frac{1}{e}\times \frac{1}{2}+\frac{1}{4}\right)$

$\Rightarrow I_2=52+\frac{52}{e}$

So, $I_1+I_2=\frac{52}{e}$ or ${\int }_{-3}^{101}\left(\left[\sin \pi x\right]+e^{\left[\cos 2\pi x\right]}\right)dx=\frac{52}{e}$