What is the integral from 1 to 3 of xe^{-5x} ?

The integral from 1 to 3 of xe^{-5x} is -(-6e^{10}+16)/(25e^{15)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 3 of xe^{-5x}

\int_{1}^{3} x e^{- 5 x} d x

- \frac{- 6 e ^{10} + 16}{25 e ^{15}}

Decimal Notation

0.00161 \dots

Solution steps

\int_{1}^{3} x e^{- 5 x} d x

Apply u-substitution

: \int_{- 5}^{- 15} \frac{e ^{u} u}{25} d u

= \int_{- 5}^{- 15} \frac{e ^{u} u}{25} d u

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

= - \int_{- 15}^{- 5} \frac{e ^{u} u}{25} d u

Take the constant out:

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

= - \frac{1}{25} \cdot \int_{- 15}^{- 5} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 15}^{- 5}

= - \frac{1}{25} [e^{u} u - \int e^{u} d u]_{- 15}^{- 5}

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

= - \frac{1}{25} [e^{u} u - e^{u}]_{- 15}^{- 5}

Compute the boundaries:

- \frac{6}{e ^{5}} + \frac{16}{e ^{15}}

= - \frac{1}{25} (- \frac{6}{e ^{5}} + \frac{16}{e ^{15}})

Simplify

= - \frac{- 6 e ^{10} + 16}{25 e ^{15}}

\int_{2}^{2} (x^{3} + 1) d x

\int_{- 1}^{2} (x + 1)^{4} d x

\int_{1}^{x} (2 - t) d t

\int_{1}^{2} x (1 + x^{2})^{3} d x

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

What is the integral from 1 to 3 of xe^{-5x} ?
The integral from 1 to 3 of xe^{-5x} is -(-6e^{10}+16)/(25e^{15)}