What is the integral from 2 to infinity of xe^{-4x} ?

The integral from 2 to infinity of xe^{-4x} is 9/(16e^8)

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integral from 2 to infinity of xe^{-4x}

\int_{2}^{\infty} x e^{- 4 x} d x

\frac{9}{16 e ^{8}}

Decimal Notation

0.00018 \dots

Solution steps

\int_{2}^{\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_{2}^{\infty} x e^{- 4 x} d x = 0 - (- \frac{9}{16 e ^{8}})

= 0 - (- \frac{9}{16 e ^{8}})

Simplify

= \frac{9}{16 e ^{8}}

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What is the integral from 2 to infinity of xe^{-4x} ?
The integral from 2 to infinity of xe^{-4x} is 9/(16e^8)