Mathematics

# Compute the integral$\displaystyle \int_{0}^{1}\left ( e^{x}-1 \right )^{4}e^{x}dx$

$\displaystyle 0.2\left ( e-1 \right )^{5}$

##### SOLUTION
Let $I=\int _{ 0 }^{ 1 }{ { \left( { e }^{ x }-1 \right) }^{ 4 } } { e }^{ x }dx\quad$

Put ${ e }^{ x }-1=t\Rightarrow { e }^{ x }dx=dt$

$x=0\implies t=0$ and $x=1 \implies t=e-1$

$\displaystyle I=\int _{ 0 }^{ e-1 }{ { t }^{ 4 } } dt=\left[ \frac { { t }^{ 5 } }{ 5 } \right] _{ 0 }^{ e-1 }{ = }\quad 0.2{ \left( e-1 \right) }^{ 5 }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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