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

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

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

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

1

Solution steps

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

Compute the indefinite integral:

\int x^{0} e^{- x} d x = - e^{- x} + C

Compute the boundaries:

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

= 0 - (- 1)

Simplify

= 1

\frac{\partial}{\partial x} (x y (4 - y))

\int_{0}^{4} 2 t e^{- t} d t

(x^{2} - 1) (\frac{d y}{d x}) + 2 y = (x + 1)^{2}

\frac{\partial}{\partial x} (\frac{y}{x} ln (x))

n = 1 \sum \infty (cos (2.6))^{n}

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