What is the integral from 0 to infinity of x^1e^{-x} ?

The integral from 0 to infinity of x^1e^{-x} is 1

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integral from 0 to infinity of x^1e^{-x}

\int_{0}^{\infty} x^{1} e^{- x} d x

1

Solution steps

\int_{0}^{\infty} x^{1} e^{- x} d x

Compute the indefinite integral:

\int x^{1} e^{- x} d x = - e^{- x} x - e^{- x} + C

Compute the boundaries:

\int_{0}^{\infty} x^{1} e^{- x} d x = 0 - (- 1)

= 0 - (- 1)

Simplify

= 1

\frac{\partial}{\partial y} (3 x^{2} y + y^{2})

\frac{d}{d x} (ln ((5 x^{2} + 9)^{3}))

\int (sin (2 x) - cos (2 x))^{2} d x

4 x \frac{d y}{d x} + y = x^{3}

\frac{\partial}{\partial x} (e^{6 x e^{y}})

What is the integral from 0 to infinity of x^1e^{-x} ?
The integral from 0 to infinity of x^1e^{-x} is 1