Question

$\underset{0}{\overset{1}{\int }}x{e}^{-5x}dx$ is equal to

• A. $\frac{1}{25}-\frac{6e^{-5}}{25}$
• B. $\frac{1}{25}+\frac{6e^{-5}}{25}$
• C. $-\frac{1}{25}-\frac{6e^{-5}}{25}$
• D. $\frac{1}{25}-\frac{1}{5}{e}^{-5}$
• E. $\frac{1}{25}+\frac{1}{5}{e}^{-5}$

Answer 1

Answer: (A)

${\int }_0^1\underset{I}{x}\underset{II}{e^{-5x}}dx$
$={\left[\left\{x\left(\frac{e^{-5x}}{-5}\right)\right\}\right]}_0^1-\left\{{\int }_0^{1}1\cdot \frac{e^{-5x}}{-5}dx\right\}$
$={\left[-\frac{x{e}^{-5x}}{5}-\frac{e^{-5x}}{25}\right]}_0^1$
$=-\frac{e^{-5}}{5}-\frac{e^{-5}}{25}+\frac{1}{25}$
$=-\frac{6e^{-5}}{25}+\frac{1}{25}=\frac{1}{25}-\frac{6e^{-5}}{25}$