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

The integral from 1 to infinity of xe^{-5x} is 6/(25e^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{6}{25 e ^{5}}

Decimal

0.00161 \dots

Solution steps

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

Compute the indefinite integral:

\int x e^{- 5 x} d x = \frac{1}{25} (- 5 e^{- 5 x} x - e^{- 5 x}) + C

Compute the boundaries:

\int_{1}^{\infty} x e^{- 5 x} d x = 0 - (- \frac{6}{25 e ^{5}})

= 0 - (- \frac{6}{25 e ^{5}})

Simplify

= \frac{6}{25 e ^{5}}

What is the integral from 1 to infinity of xe^{-5x} ?
The integral from 1 to infinity of xe^{-5x} is 6/(25e^5)