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

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

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

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

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

Decimal Notation

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}}

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