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

The integral from 5 to infinity of xe^{-5x} is (26)/(25e^{25)}

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

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

\frac{26}{25 e ^{25}}

Decimal

1.44435 E - 11

Solution steps

\int_{5}^{\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_{5}^{\infty} x e^{- 5 x} d x = 0 - (- \frac{26}{25 e ^{25}})

= 0 - (- \frac{26}{25 e ^{25}})

Simplify

= \frac{26}{25 e ^{25}}

\int_{0}^{\frac{1}{2}} e^{- x} d x

\int_{1}^{2} \frac{1}{x ^{2} + 2 x + 5} d x

\int_{0}^{10} 2 e^{- 2 x} d x

\int_{0}^{2} 20 t^{2} e^{- t} d t

\int_{0}^{1} y e^{- 2 y^{2}} d y

What is the integral from 5 to infinity of xe^{-5x} ?
The integral from 5 to infinity of xe^{-5x} is (26)/(25e^{25)}