What is the integral from 8 to infinity of xe^{-5x} ?

The integral from 8 to infinity of xe^{-5x} is (41)/(25e^{40)}

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integral from 8 to infinity of xe^{-5x}

\int_{8}^{\infty} x e^{- 5 x} d x

\frac{41}{25 e ^{40}}

Decimal Notation

6.9673 E - 18

Solution steps

\int_{8}^{\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_{8}^{\infty} x e^{- 5 x} d x = 0 - (- \frac{41}{25 e ^{40}})

= 0 - (- \frac{41}{25 e ^{40}})

Simplify

= \frac{41}{25 e ^{40}}

\int \frac{( ln ( x ) ) ^{12}}{x} d x

\frac{d}{d x} (12 e^{\frac{x}{2}})

\int \frac{e ^{x} ( x ln ( x ) + 1 )}{x} d x

\frac{3 x + 8}{x ^{2}}

\frac{( 3 s + 1 )}{s + 2}

What is the integral from 8 to infinity of xe^{-5x} ?
The integral from 8 to infinity of xe^{-5x} is (41)/(25e^{40)}