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

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

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

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

- \frac{65}{e} + 24

Decimal Notation

0.08783 \dots

Solution steps

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

Apply Integration By Parts

: [- e^{- x} x^{4} - \int - 4 e^{- x} x^{3} d x]_{0}^{1}

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

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

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

Simplify

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

Compute the boundaries:

- \frac{65}{e} + 24

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

n = 0 \sum \infty 5 \cdot 3^{n}

f (x) = (x^{2} - 1) (x^{2} + 1)

n = 1 \sum \infty \frac{x ^{n}}{1 1 ^{n}}

\int 2 (x - 2) d x

\int cos (t - x) d x

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