What is the integral from 1 to infinity of xe^{-4x} ?

The integral from 1 to infinity of xe^{-4x} is 5/(16e^4)

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integral from 1 to infinity of xe^{-4x}

\int_{1}^{\infty} x e^{- 4 x} d x

\frac{5}{16 e ^{4}}

Solution steps

\int_{1}^{\infty} x e^{- 4 x} d x

Compute the indefinite integral:

\int x e^{- 4 x} d x = \frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Compute the boundaries:

\int_{1}^{\infty} x e^{- 4 x} d x = 0 - (- \frac{5}{16 e ^{4}})

= 0 - (- \frac{5}{16 e ^{4}})

Simplify

= \frac{5}{16 e ^{4}}

\int sin^{4} (4 θ) d θ

\int (\frac{16}{1 + x ^{2}}) d x

sin (x^{2})

\frac{d}{d x} (x \cdot e^{4 x})

\int \frac{( x + 2 )}{x ^{2} + 3 x - 4} d x

What is the integral from 1 to infinity of xe^{-4x} ?
The integral from 1 to infinity of xe^{-4x} is 5/(16e^4)