What is the integral from 6 to infinity of xe^{-3x} ?

The integral from 6 to infinity of xe^{-3x} is (19)/(9e^{18)}

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integral from 6 to infinity of xe^{-3x}

\int_{6}^{\infty} x e^{- 3 x} d x

\frac{19}{9 e ^{18}}

Decimal Notation

3.21522 E - 8

Solution steps

\int_{6}^{\infty} x e^{- 3 x} d x

Compute the indefinite integral:

\int x e^{- 3 x} d x = \frac{1}{9} (- 3 e^{- 3 x} x - e^{- 3 x}) + C

Compute the boundaries:

\int_{6}^{\infty} x e^{- 3 x} d x = 0 - (- \frac{19}{9 e ^{18}})

= 0 - (- \frac{19}{9 e ^{18}})

Simplify

= \frac{19}{9 e ^{18}}

\frac{d}{d x} (\frac{1}{3} x^{3} - x^{2} - 3 x + 4)

7 x^{4} - 7 x^{2}, 8 x^{2}, 0

\frac{\partial}{\partial y} (e^{- (x^{2} + y^{2})})

x \to 4 lim (\frac{( ln ( x ) - ln ( 4 ) )}{x - 4})

y^{^{'}} = x^{2} y^{2}

What is the integral from 6 to infinity of xe^{-3x} ?
The integral from 6 to infinity of xe^{-3x} is (19)/(9e^{18)}