What is the integral from 0 to 1 of (xe^{-2x})/2 ?

The integral from 0 to 1 of (xe^{-2x})/2 is -(-e^2+3)/(8e^2)

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integral from 0 to 1 of (xe^{-2x})/2

\int_{0}^{1} \frac{x e ^{- 2 x}}{2} d x

- \frac{- e ^{2} + 3}{8 e ^{2}}

Decimal Notation

0.07424 \dots

Solution steps

\int_{0}^{1} \frac{x e ^{- 2 x}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} x e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 2} \frac{e ^{u} u}{4} d u

= \frac{1}{2} \cdot \int_{0}^{- 2} \frac{e ^{u} u}{4} d u

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

= \frac{1}{2} (- \int_{- 2}^{0} \frac{e ^{u} u}{4} d u)

Take the constant out:

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

= \frac{1}{2} (- \frac{1}{4} \cdot \int_{- 2}^{0} e^{u} u d u)

Apply Integration By Parts

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

= \frac{1}{2} (- \frac{1}{4} [e^{u} u - \int e^{u} d u]_{- 2}^{0})

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

= \frac{1}{2} (- \frac{1}{4} [e^{u} u - e^{u}]_{- 2}^{0})

Simplify

\frac{1}{2} (- \frac{1}{4} [e^{u} u - e^{u}]_{- 2}^{0}) : - \frac{1}{8} [e^{u} u - e^{u}]_{- 2}^{0}

= - \frac{1}{8} [e^{u} u - e^{u}]_{- 2}^{0}

Compute the boundaries:

- 1 + \frac{3}{e ^{2}}

= - \frac{1}{8} (- 1 + \frac{3}{e ^{2}})

Simplify

= - \frac{- e ^{2} + 3}{8 e ^{2}}

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What is the integral from 0 to 1 of (xe^{-2x})/2 ?
The integral from 0 to 1 of (xe^{-2x})/2 is -(-e^2+3)/(8e^2)