What is the integral from 0 to 1 of xe^{-9x} ?

The integral from 0 to 1 of xe^{-9x} is -(-e^9+10)/(81e^9)

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integral from 0 to 1 of xe^{-9x}

\int_{0}^{1} x e^{- 9 x} d x

- \frac{- e ^{9} + 10}{81 e ^{9}}

Decimal

0.01233 \dots

Solution steps

\int_{0}^{1} x e^{- 9 x} d x

Apply u-substitution

: \int_{0}^{- 9} \frac{e ^{u} u}{81} d u

= \int_{0}^{- 9} \frac{e ^{u} u}{81} d u

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

= - \int_{- 9}^{0} \frac{e ^{u} u}{81} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{81} \cdot \int_{- 9}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 9}^{0}

= - \frac{1}{81} [e^{u} u - \int e^{u} d u]_{- 9}^{0}

\int e^{u} d u = e^{u}

= - \frac{1}{81} [e^{u} u - e^{u}]_{- 9}^{0}

Compute the boundaries:

- 1 + \frac{10}{e ^{9}}

= - \frac{1}{81} (- 1 + \frac{10}{e ^{9}})

Simplify

= - \frac{- e ^{9} + 10}{81 e ^{9}}

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What is the integral from 0 to 1 of xe^{-9x} ?
The integral from 0 to 1 of xe^{-9x} is -(-e^9+10)/(81e^9)