Question

If the integral ${\int }_0^{10}\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{x-\left[x\right]}}dx=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma$, where $\alpha ,\beta ,\gamma$ are integers and $\left[x\right]$ denotes the greatest integer less than or equal to $x$, then the value of $\alpha +\beta +\gamma$ is equal to.

• A. $20$
• B. $0$
• C. $25$
• D. $10$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

The given integration, $I={\int }_0^{10}\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{x-\left[x\right]}}dx$.

It is known that, $x=\left[x\right]+\left\{x\right\}$, where $\left\{x\right\}$ is the fractional part function.

$\Rightarrow x-\left[x\right]=\left\{x\right\}$.

Thus, $I={\int }_0^{10}\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{\left\{x\right\}}}dx$.

As $\left\{x\right\}$ and $\left[\sin \left(2\mathrm{\pi x}\right)\right]$ both are periodic with the period of $1$, thus $\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{\left\{x\right\}}}$ also have a period of $1$.

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

Now, the value of $\sin \left(\theta \right)$ is in the interval $\left[0,1\right]$, when $\theta \in \left[0,\mathrm{\pi }\right]$ and is in the interval $\left[-1,0\right]$, when $\theta \in \left[\mathrm{\pi },2\mathrm{\pi }\right]$.

Thus, $\left[\sin \left(\theta \right)\right]=0$, when $\theta \in \left[0,\mathrm{\pi }\right]$ and $\left[\sin \left(\theta \right)\right]=-1$, when $\theta \in \left[\mathrm{\pi },2\mathrm{\pi }\right]$.

$$\begin{gathered} I=10{\int }_0^1\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{\left\{x\right\}}}dx \\ \Rightarrow I=10{\int }_0^{\frac{1}{2}}\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{\left\{x\right\}}}dx+10{\int }_{\frac{1}{2}}^1\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{\left\{x\right\}}}dx \\ \Rightarrow I=10{\int }_0^{\frac{1}{2}}\frac{0}{e^{\left\{x\right\}}}dx+10{\int }_{\frac{1}{2}}^1\frac{-1}{e^{\left\{x\right\}}}dx \\ \Rightarrow I=-10{\int }_{\frac{1}{2}}^1{e}^{-x}dx \\ \Rightarrow I=-10{\left[-e^{-x}\right]}_{\frac{1}{2}}^1 \\ \Rightarrow I=10e^{-1}-10e^{-\frac{1}{2}} \end{gathered}$$

It is given that ${\int }_0^{10}\frac{\left[\sin \left(2\mathrm{\pi x}\right)\right]}{e^{x-\left[x\right]}}dx=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma$

$\Rightarrow 10e^{-1}-10e^{-\frac{1}{2}}=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma$

Therefore, $\alpha =10$, $\beta =-10$ and $\gamma =0$.

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

$\Rightarrow \alpha +\beta +\gamma =0$.

Hence, option (B) is the correct answer.