What is the integral from 0 to infinity of x^2e^{-x} ?

The integral from 0 to infinity of x^2e^{-x} is 2

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integral from 0 to infinity of x^2e^{-x}

\int_{0}^{\infty} x^{2} e^{- x} d x

2

Solution steps

\int_{0}^{\infty} x^{2} e^{- x} d x

Compute the indefinite integral:

\int x^{2} e^{- x} d x = - e^{- x} x^{2} + 2 (- e^{- x} x - e^{- x}) + C

Compute the boundaries:

\int_{0}^{\infty} x^{2} e^{- x} d x = 0 - (- 2)

= 0 - (- 2)

Simplify

= 2

(3.5) (2.6)

\frac{d}{d x} (g (x) (x)^{2} (x) + x)

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

x \to 0 lim (x^{6} cos (\frac{9}{x}))

x \to - \infty lim (5 x^{3})

What is the integral from 0 to infinity of x^2e^{-x} ?
The integral from 0 to infinity of x^2e^{-x} is 2