What is the integral from 0 to 1 of xe^{-8x} ?

The integral from 0 to 1 of xe^{-8x} is -(-e^8+9)/(64e^8)

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integral from 0 to 1 of xe^{-8x}

\int_{0}^{1} x e^{- 8 x} d x

- \frac{- e ^{8} + 9}{64 e ^{8}}

Decimal

0.01557 \dots

Solution steps

\int_{0}^{1} x e^{- 8 x} d x

Apply u-substitution

: \int_{0}^{- 8} \frac{e ^{u} u}{64} d u

= \int_{0}^{- 8} \frac{e ^{u} u}{64} d u

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

= - \int_{- 8}^{0} \frac{e ^{u} u}{64} d u

Take the constant out:

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

= - \frac{1}{64} \cdot \int_{- 8}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 8}^{0}

= - \frac{1}{64} [e^{u} u - \int e^{u} d u]_{- 8}^{0}

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

= - \frac{1}{64} [e^{u} u - e^{u}]_{- 8}^{0}

Compute the boundaries:

- 1 + \frac{9}{e ^{8}}

= - \frac{1}{64} (- 1 + \frac{9}{e ^{8}})

Simplify

= - \frac{- e ^{8} + 9}{64 e ^{8}}

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What is the integral from 0 to 1 of xe^{-8x} ?
The integral from 0 to 1 of xe^{-8x} is -(-e^8+9)/(64e^8)