What is the integral from 0 to 3 of xe^{-2x} ?

The integral from 0 to 3 of xe^{-2x} is -(-e^6+7)/(4e^6)

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integral from 0 to 3 of xe^{-2x}

\int_{0}^{3} x e^{- 2 x} d x

- \frac{- e ^{6} + 7}{4 e ^{6}}

Decimal Notation

0.24566 \dots

Solution steps

\int_{0}^{3} x e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 6} \frac{e ^{u} u}{4} d u

= \int_{0}^{- 6} \frac{e ^{u} u}{4} d u

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

= - \int_{- 6}^{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}{4} \cdot \int_{- 6}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 6}^{0}

= - \frac{1}{4} [e^{u} u - \int e^{u} d u]_{- 6}^{0}

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

= - \frac{1}{4} [e^{u} u - e^{u}]_{- 6}^{0}

Compute the boundaries:

- 1 + \frac{7}{e ^{6}}

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

Simplify

= - \frac{- e ^{6} + 7}{4 e ^{6}}

\int \frac{x ^{2}}{9} d x

f (x) = 5 e^{x} + x, (0, 5)

x \to b lim (\frac{x - b}{x - b})

\int \frac{x + 21}{x ^{2} + 12 x + 40} d x

n = 0 \sum \infty \frac{n ^{3}}{3 ^{n}}

What is the integral from 0 to 3 of xe^{-2x} ?
The integral from 0 to 3 of xe^{-2x} is -(-e^6+7)/(4e^6)