What is the integral from 2 to infinity of xe^{-8x} ?

The integral from 2 to infinity of xe^{-8x} is (17)/(64e^{16)}

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integral from 2 to infinity of xe^{-8x}

\int_{2}^{\infty} x e^{- 8 x} d x

\frac{17}{64 e ^{16}}

Solution steps

\int_{2}^{\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_{2}^{\infty} x e^{- 8 x} d x = 0 - (- \frac{17}{64 e ^{16}})

= 0 - (- \frac{17}{64 e ^{16}})

Simplify

= \frac{17}{64 e ^{16}}

e^{x}, x e^{x}, 0, 1

y = \frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}

\int - \frac{1}{2} cos (2 x) d x

\frac{( 2 s + 1 )}{( s ^{2} - 2 s + 2 )}

\frac{\partial}{\partial x} ((y^{2}) d x)

What is the integral from 2 to infinity of xe^{-8x} ?
The integral from 2 to infinity of xe^{-8x} is (17)/(64e^{16)}