What is the integral from 3 to 7 of xe^{-2x} ?

The integral from 3 to 7 of xe^{-2x} is -(-7e^8+15)/(4e^{14)}

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integral from 3 to 7 of xe^{-2x}

\int_{3}^{7} x e^{- 2 x} d x

- \frac{- 7 e ^{8} + 15}{4 e ^{14}}

Decimal Notation

0.00433 \dots

Solution steps

\int_{3}^{7} x e^{- 2 x} d x

Apply u-substitution

: \int_{- 6}^{- 14} \frac{e ^{u} u}{4} d u

= \int_{- 6}^{- 14} \frac{e ^{u} u}{4} d u

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

= - \int_{- 14}^{- 6} \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}{4} \cdot \int_{- 14}^{- 6} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 14}^{- 6}

= - \frac{1}{4} [e^{u} u - \int e^{u} d u]_{- 14}^{- 6}

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

= - \frac{1}{4} [e^{u} u - e^{u}]_{- 14}^{- 6}

Compute the boundaries:

- \frac{7}{e ^{6}} + \frac{15}{e ^{14}}

= - \frac{1}{4} (- \frac{7}{e ^{6}} + \frac{15}{e ^{14}})

Simplify

= - \frac{- 7 e ^{8} + 15}{4 e ^{14}}

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What is the integral from 3 to 7 of xe^{-2x} ?
The integral from 3 to 7 of xe^{-2x} is -(-7e^8+15)/(4e^{14)}