What is the integral from 7 to infinity of xe^{-5x} ?

The integral from 7 to infinity of xe^{-5x} is (36)/(25e^{35)}

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integral from 7 to infinity of xe^{-5x}

\int_{7}^{\infty} x e^{- 5 x} d x

\frac{36}{25 e ^{35}}

Decimal Notation

9.07937 E - 16

Solution steps

\int_{7}^{\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_{7}^{\infty} x e^{- 5 x} d x = 0 - (- \frac{36}{25 e ^{35}})

= 0 - (- \frac{36}{25 e ^{35}})

Simplify

= \frac{36}{25 e ^{35}}

\frac{d}{d x} (4 x^{3} - 102 x^{2} + 840 x + 520)

y^{^{''}} + 5 y^{^{'}} + 6 y = 1 + 2 t + 3 t^{2}

25 (x^{2} - y^{2})

tan (3 x)

\frac{5 x}{x + 3}

What is the integral from 7 to infinity of xe^{-5x} ?
The integral from 7 to infinity of xe^{-5x} is (36)/(25e^{35)}