What is the integral from 7 to infinity of xe^{-4x} ?

The integral from 7 to infinity of xe^{-4x} is (29)/(16e^{28)}

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integral from 7 to infinity of xe^{-4x}

\int_{7}^{\infty} x e^{- 4 x} d x

\frac{29}{16 e ^{28}}

Decimal Notation

1.25324 E - 12

Solution steps

\int_{7}^{\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_{7}^{\infty} x e^{- 4 x} d x = 0 - (- \frac{29}{16 e ^{28}})

= 0 - (- \frac{29}{16 e ^{28}})

Simplify

= \frac{29}{16 e ^{28}}

\frac{\partial}{\partial x} (5 (y - 1)^{3} (x - 4))

x \to \infty + lim (\frac{2}{x} - 3)

\frac{\partial}{\partial x} (y - 2 x - 2)

\int \frac{2}{2 t - 1} d t

What is the integral from 7 to infinity of xe^{-4x} ?
The integral from 7 to infinity of xe^{-4x} is (29)/(16e^{28)}