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Question

Mathematics Question on Integrals of Some Particular Functions

∫01xe−5x dx\int\limits_{0}^{1} x e^{-5x} \, dx is equal to

A

125−6e−525\frac{1}{25}-\frac{6e^{-5}}{25}

B

125+6e−525\frac{1}{25}+\frac{6e^{-5}}{25}

C

−125−6e−525-\frac{1}{25}-\frac{6e^{-5}}{25}

D

125−15e−5\frac{1}{25}-\frac{1}{5}e^{-5}

Answer

125−6e−525\frac{1}{25}-\frac{6e^{-5}}{25}

Explanation

Solution

∫01 xI e−5xII dx\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\\}
=[−xe−5x5−e−5x25]01=\left[-\frac{x e^{-5 x}}{5}-\frac{e^{-5 x}}{25}\right]_{0}^{1}
=−e−55−e−525+125=-\frac{e^{-5}}{5}-\frac{e^{-5}}{25}+\frac{1}{25}
=−6e−525+125=125−6e−525=-\frac{6 e^{-5}}{25}+\frac{1}{25}=\frac{1}{25}-\frac{6 e^{-5}}{25}