What is the integral from 0 to 1 of p^4e^{-p} ?

The integral from 0 to 1 of p^4e^{-p} is -(65)/e+24

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integral from 0 to 1 of p^4e^{-p}

\int_{0}^{1} p^{4} e^{- p} d p

- \frac{65}{e} + 24

Decimal

0.08783 \dots

Solution steps

\int_{0}^{1} p^{4} e^{- p} d p

Apply Integration By Parts

= [- e^{- p} p^{4} - \int - 4 e^{- p} p^{3} d p]_{0}^{1}

\int - 4 e^{- p} p^{3} d p = - 4 (- e^{- p} p^{3} - 3 (e^{- p} p^{2} - 2 (- e^{- p} p - e^{- p})))

= [- e^{- p} p^{4} - (- 4 (- e^{- p} p^{3} - 3 (e^{- p} p^{2} - 2 (- e^{- p} p - e^{- p}))))]_{0}^{1}

Simplify

= [- e^{- p} p^{4} + 4 (- e^{- p} p^{3} - 3 (e^{- p} p^{2} - 2 (- e^{- p} p - e^{- p})))]_{0}^{1}

Compute the boundaries:

- \frac{65}{e} + 24

= - \frac{65}{e} + 24

What is the integral from 0 to 1 of p^4e^{-p} ?
The integral from 0 to 1 of p^4e^{-p} is -(65)/e+24