What is the integral from 0 to 2 of xe^{-5x} ?

The integral from 0 to 2 of xe^{-5x} is -(-e^{10}+11)/(25e^{10)}

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integral from 0 to 2 of xe^{-5x}

\int_{0}^{2} x e^{- 5 x} d x

- \frac{- e ^{10} + 11}{25 e ^{10}}

Decimal

0.03998 \dots

Solution steps

\int_{0}^{2} x e^{- 5 x} d x

Apply u-substitution

: \int_{0}^{- 10} \frac{e ^{u} u}{25} d u

= \int_{0}^{- 10} \frac{e ^{u} u}{25} d u

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

= - \int_{- 10}^{0} \frac{e ^{u} u}{25} d u

Take the constant out:

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

= - \frac{1}{25} \cdot \int_{- 10}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 10}^{0}

= - \frac{1}{25} [e^{u} u - \int e^{u} d u]_{- 10}^{0}

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

= - \frac{1}{25} [e^{u} u - e^{u}]_{- 10}^{0}

Compute the boundaries:

- 1 + \frac{11}{e ^{10}}

= - \frac{1}{25} (- 1 + \frac{11}{e ^{10}})

Simplify

= - \frac{- e ^{10} + 11}{25 e ^{10}}

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What is the integral from 0 to 2 of xe^{-5x} ?
The integral from 0 to 2 of xe^{-5x} is -(-e^{10}+11)/(25e^{10)}