Question

${\int }_{-\frac{1}{7}}^{\frac{20}{7}}{e}^{5\left\{x\right\}}dx$ where $\left\{\right\}=$ fractional part of $x$

• A. $\frac{3}{5}\left(e^5\right)$
• B. $\frac{3}{5}(e^5-1)$
• C. $\frac{3}{5}$
• D. $0$

Answer 1

Answer: (D)

$0$

Let $I={\int }_{\frac{-1}{7}}^{\frac{20}{7}}{e}^{5\left\{x\right\}}$

$\left\{x\right\}=$ fractional part of $x$.ie. for example

$\left\{3.64\right\}=3.64-\left[3.64\right]=3.64-3=0.64$

$\{-3.64\}=-3.64-[-3.64]=-3.64-(-4)=0.36$

$\left\{x\right\}$ is non-negative.

$I={\int }_{\frac{-1}{7}}^{\frac{20}{7}}{e}^{5\left\{x\right\}}$

$=\frac{e^{5\left\{x\right\}}}{5}{|}_{\frac{-1}{7}}^{\frac{20}{7}}$

$=\frac{1}{5}[e^{5\left(\frac{6}{7}\right)}-e^{5\left(\frac{6}{7}\right)}]$

$=0$ ..... $[\because \left\{\frac{20}{7}\right\}=\frac{20}{7}-\left[\frac{20}{7}\right]=2+\frac{6}{7}-2=\frac{6}{7}\text{and}\left\{\frac{-1}{7}\right\}=\frac{-1}{7}+1=\frac{6}{7}]$