What is the integral from 0 to 1 of x^3e^{-x} ?

The integral from 0 to 1 of x^3e^{-x} is 6-(16)/e

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integral from 0 to 1 of x^3e^{-x}

\int_{0}^{1} x^{3} e^{- x} d x

6 - \frac{16}{e}

Decimal

0.11392 \dots

Solution steps

\int_{0}^{1} x^{3} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 1} e^{u} u^{3} d u

= \int_{0}^{- 1} e^{u} u^{3} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} e^{u} u^{3} d u

Apply Integration By Parts

: [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- 1}^{0}

= - [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- 1}^{0}

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

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

Compute the boundaries:

- 6 + \frac{16}{e}

= - (- 6 + \frac{16}{e})

Simplify

= 6 - \frac{16}{e}

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What is the integral from 0 to 1 of x^3e^{-x} ?
The integral from 0 to 1 of x^3e^{-x} is 6-(16)/e