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

The integral from 0 to 2 of 2xe^{-3x} is -(2(-e^6+7))/(9e^6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{2 ( - e ^{6} + 7 )}{9 e ^{6}}

Decimal Notation

0.21836 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{- 6} \frac{e ^{u} u}{9} d u

= 2 \cdot \int_{0}^{- 6} \frac{e ^{u} u}{9} d u

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

= 2 (- \int_{- 6}^{0} \frac{e ^{u} u}{9} d u)

Take the constant out:

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

= 2 (- \frac{1}{9} \cdot \int_{- 6}^{0} e^{u} u d u)

Apply Integration By Parts

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

= 2 (- \frac{1}{9} [e^{u} u - \int e^{u} d u]_{- 6}^{0})

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

= 2 (- \frac{1}{9} [e^{u} u - e^{u}]_{- 6}^{0})

Simplify

2 (- \frac{1}{9} [e^{u} u - e^{u}]_{- 6}^{0}) : - \frac{2}{9} [e^{u} u - e^{u}]_{- 6}^{0}

= - \frac{2}{9} [e^{u} u - e^{u}]_{- 6}^{0}

Compute the boundaries:

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

= - \frac{2}{9} (- 1 + \frac{7}{e ^{6}})

Simplify

= - \frac{2 ( - e ^{6} + 7 )}{9 e ^{6}}

\int_{0}^{\frac{π}{4}} 8 sec (θ) tan (θ) d θ

\int_{0}^{π} sin^{3} (\frac{t}{2}) d t

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

\int_{- \frac{π}{3}}^{\frac{π}{4}} cos (θ) d θ

\int_{3}^{\infty} \frac{7}{x ^{2} + 2} d x

What is the integral from 0 to 2 of 2xe^{-3x} ?
The integral from 0 to 2 of 2xe^{-3x} is -(2(-e^6+7))/(9e^6)