What is the integral from 1 to infinity of xe^{-8x} ?

The integral from 1 to infinity of xe^{-8x} is 9/(64e^8)

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integral from 1 to infinity of xe^{-8x}

\int_{1}^{\infty} x e^{- 8 x} d x

\frac{9}{64 e ^{8}}

Decimal Notation

0.00004 \dots

Solution steps

\int_{1}^{\infty} x e^{- 8 x} d x

Compute the indefinite integral:

\int x e^{- 8 x} d x = \frac{1}{64} (- 8 e^{- 8 x} x - e^{- 8 x}) + C

Compute the boundaries:

\int_{1}^{\infty} x e^{- 8 x} d x = 0 - (- \frac{9}{64 e ^{8}})

= 0 - (- \frac{9}{64 e ^{8}})

Simplify

= \frac{9}{64 e ^{8}}

\frac{d}{d x} ((4 x - 1)^{2})

y = x^{2}, y = 2 x^{2} - 4 x, y = 0

\int \frac{x ^{2}}{( x ^{2} + 8 ) ^{\frac{3}{2}}} d x

\frac{d r}{d θ} = π

x^{3} - 8 x

What is the integral from 1 to infinity of xe^{-8x} ?
The integral from 1 to infinity of xe^{-8x} is 9/(64e^8)