Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_0^{10}\frac{\left[\sin 2\mathrm{\pi }x\right]}{e^{x-\left[x\right]}}dx={\int }_0^{10}\frac{\left[\sin 2\mathrm{\pi }x\right]}{e^{\left\{x\right\}}}dx,$ where $x-\left[x\right]=\left\{x\right\}$ is the fractional part function.

We know that the $\left\{x\right\}$ is a periodic function with period $1$ and also that $\sin 2\pi x$ is a periodic function with period $\frac{2\mathrm{\pi }}{2\mathrm{\pi }}=1.$

Thus, the function $f\left(x\right)=\frac{\left[\sin 2\mathrm{\pi }x\right]}{e^{\left\{x\right\}}}$ is periodic with period $1.$

If $f\left(x\right)$ is a periodic function with period $T$ then ${\int }_0^{nT}f\left(x\right)dx=n{\int }_0^{T}f\left(x\right)dx, n\in N$

Therefore $I=10{\int }_0^1\frac{\left[\sin 2\mathrm{\pi }x\right]}{e^{\left\{x\right\}}}dx$

$\Rightarrow I=10{\int }_0^1\frac{\left[\sin 2\mathrm{\pi }x\right]}{e^x}dx$

Now, for $0\le x\le 1 \Rightarrow -1\le \sin 2\mathrm{\pi x}\le 1,$ hence $\left[\sin 2x\right]$ can take values $0$ and $-1$

$\Rightarrow I=10\left({\int }_0^{1/2}\frac{\left[\sin 2\mathrm{\pi }x\right]}{e^x}dx+{\int }_{1/2}^1\frac{\left[\sin 2\mathrm{\pi }x\right]}{e^x}dx\right)$

$\Rightarrow I=10\left(0+{\int }_{1/2}^1\frac{(-1)}{e^x}dx\right)$

$\Rightarrow I=-10{\int }_{1/2}^1{e}^{-x}dx$

$\Rightarrow I=10{\left[e^{-x}\right]}_{1/2}^1$

$\Rightarrow I=10\left({\mathrm{e}}^{-1}-{\mathrm{e}}^{-1/2}\right)$

Given, $I=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma ,$ thus $\alpha =10, \beta =-10, \gamma =0.$

And, $\alpha +\beta +\gamma =10-10=0.$