Question

$\int\limits_{0}^{1} x e^{-5x} \, dx$ 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)

$\int_{0}^{1} \,\underset{I}{x}\, \underset{II}{e^{-5x}}\,dx$
$=\left[\left\{x\left(\frac{e^{-5 x}}{-5}\right)\right\}\right]_{0}^{1}-\left\{\int_{0}^{1} 1 \cdot \frac{e^{-5 x}}{-5} d x\right\}$
$=\left[-\frac{x e^{-5 x}}{5}-\frac{e^{-5 x}}{25}\right]_{0}^{1}$
$=-\frac{e^{-5}}{5}-\frac{e^{-5}}{25}+\frac{1}{25}$
$=-\frac{6 e^{-5}}{25}+\frac{1}{25}=\frac{1}{25}-\frac{6 e^{-5}}{25}$