What is the integral from 0 to 1 of x^7e^{-x} ?

The integral from 0 to 1 of x^7e^{-x} is 5040-(13700)/e

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integral from 0 to 1 of x^7e^{-x}

\int_{0}^{1} x^{7} e^{- x} d x

5040 - \frac{13700}{e}

Decimal Notation

0.05165 \dots

Solution steps

\int_{0}^{1} x^{7} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 1} e^{u} u^{7} d u

= \int_{0}^{- 1} e^{u} u^{7} d u

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

= - \int_{- 1}^{0} e^{u} u^{7} d u

Apply u-substitution

: \int_{\frac{1}{e}}^{1} ln^{7} (v) d v

= - \int_{\frac{1}{e}}^{1} ln^{7} (v) d v

Apply Integration By Parts

: [v ln^{7} (v) - \int 7 ln^{6} (v) d v]_{\frac{1}{e}}^{1}

= - [v ln^{7} (v) - \int 7 ln^{6} (v) d v]_{\frac{1}{e}}^{1}

\int 7 ln^{6} (v) d v = 7 (v ln^{6} (v) - 6 (v ln^{5} (v) - 5 (v ln^{4} (v) - 4 (v ln^{3} (v) - 3 (v ln^{2} (v) - 2 (v ln (v) - v))))))

= - [v ln^{7} (v) - 7 (v ln^{6} (v) - 6 (v ln^{5} (v) - 5 (v ln^{4} (v) - 4 (v ln^{3} (v) - 3 (v ln^{2} (v) - 2 (v ln (v) - v))))))]_{\frac{1}{e}}^{1}

Compute the boundaries:

- 5040 + \frac{13700}{e}

= - (- 5040 + \frac{13700}{e})

Simplify

= 5040 - \frac{13700}{e}

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What is the integral from 0 to 1 of x^7e^{-x} ?
The integral from 0 to 1 of x^7e^{-x} is 5040-(13700)/e