Question

Evaluate the Integral \(\int\limits_0^1 {{x^4}{e^{ - 4}}dx} \)

Short answer

The value of the integral is \(24 - \frac{{65}}{e}\)

Step-by-step solution

Applying integration by parts four times

\(\int\limits_0^1 {{x^4}{e^{ - x}}dx = \left( {{x^4}\frac{{{e^{ - x}}}}{{ - 1}}} \right)} - \int\limits_0^1 {4{x^3}} \frac{{{e^{ - x}}}}{{ - 1}}dx + {C_1}\) \(\begin{aligned}{l}\left( { - {e^{ - 1}} + 0} \right) + 4\int {4{x^3}\frac{{{e^{ - x}}}}{{ - 1}}} dx + {C_1}\\ - \frac{1}{e} + 4(\left( {{x^3}\frac{{{e^{ - x}}}}{{ - 1}}} \right)_0^1 - \int\limits_0^1 {{x^3}\frac{{{e^{ - x}}}}{{ - 1}}dx} ) + {C_1}\\ - \frac{1}{e} + 4\left( { - {e^{ - 1}} + 0} \right) + 12(\left( {{x^2}\frac{{{e^{ - x}}}}{{ - 1}}} \right)_0^1 - \int\limits_0^1 {2x\frac{{{e^{ - x}}}}{{ - 1}}dx} ) + {C_3}\\\frac{{ - 5}}{e} - \frac{{12}}{e} + 24\int\limits_0^1 {x{e^{ - x}}dx + {C_3}} \end{aligned}\)

\(\begin{aligned}{l} - \frac{{17}}{e} + 24(\left( {x\frac{{{e^{ - x}}}}{{ - 1}}} \right)_0^1 - \int\limits_0^1 {\frac{{{e^{ - x}}}}{{ - 1}}dx} ) + {C_4}\\ - \frac{{17}}{e} - \frac{{24}}{e} + 24\int\limits_0^1 {{e^{ - x}}dx + {C_4}} \end{aligned}\)

Evaluate the integral

\(\begin{aligned}{l} &= \frac{{ - 41}}{e} + 24\left( {\frac{{{e^{ - x}}}}{{ - 1}}} \right)_0^1\\ &= \frac{{ - 41}}{e} + \frac{{24}}{e} + 24{e^0}\\ &= \frac{{ - 65}}{e} + 24\end{aligned}\)

Hence, \(\int\limits_0^1 {{x^4}{e^{ - 4}}dx = 24 - \frac{{65}}{e}} \)