What is the integral from 6 to infinity of xe^{-4x} ?

The integral from 6 to infinity of xe^{-4x} is (25)/(16e^{24)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 6 to infinity of xe^{-4x}

\int_{6}^{\infty} x e^{- 4 x} d x

\frac{25}{16 e ^{24}}

Decimal

5.89865 E - 11

Solution steps

\int_{6}^{\infty} x e^{- 4 x} d x

Compute the indefinite integral:

\int x e^{- 4 x} d x = \frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Compute the boundaries:

\int_{6}^{\infty} x e^{- 4 x} d x = 0 - (- \frac{25}{16 e ^{24}})

= 0 - (- \frac{25}{16 e ^{24}})

Simplify

= \frac{25}{16 e ^{24}}

What is the integral from 6 to infinity of xe^{-4x} ?
The integral from 6 to infinity of xe^{-4x} is (25)/(16e^{24)}