What is the integral from 0 to 2 of 9xe^{-3x} ?

The integral from 0 to 2 of 9xe^{-3x} is 1-7/(e^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 9xe^{-3x}

\int_{0}^{2} 9 x e^{- 3 x} d x

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

Decimal Notation

0.98264 \dots

Solution steps

\int_{0}^{2} 9 x e^{- 3 x} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{2} x e^{- 3 x} d x

Apply u-substitution

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

= 9 \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

= 9 (- \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

= 9 (- \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}

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

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

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

Simplify

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

= - [e^{u} u - e^{u}]_{- 6}^{0}

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{2} \frac{1}{9 x ^{2} + 36} d x

\int_{2}^{5} 2 x - 1 d x

\int_{0}^{1} \frac{3 y ^{2}}{4 + y ^{3}} d y

\int_{2}^{3} \frac{4 t}{1 + t ^{2}} d t

\int_{0}^{\frac{π}{2}} sin (x) cos (y) d y

What is the integral from 0 to 2 of 9xe^{-3x} ?
The integral from 0 to 2 of 9xe^{-3x} is 1-7/(e^6)