What is the integral from 0 to infinity of (e^{-9x}) ?

The integral from 0 to infinity of (e^{-9x}) is 1/9

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integral from 0 to infinity of (e^{-9x})

\int_{0}^{\infty} (e^{- 9 x}) d x

\frac{1}{9}

Decimal Notation

0.11111 \dots

Solution steps

\int_{0}^{\infty} e^{- 9 x} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{\infty} (e^{- 9 x}) d x = 0 - (- \frac{1}{9})

= 0 - (- \frac{1}{9})

Simplify

= \frac{1}{9}

x \to 0 lim (\frac{x ^{2} + 8 x}{x})

\frac{d}{d x} (arccsc (2 x))

\frac{d y}{d t} - 3 y = 3 e^{4 t}

y = 2 x^{2} lo g_{e} (x) - x^{2}

f (x) = e^{x}, at x = ln (7)

What is the integral from 0 to infinity of (e^{-9x}) ?
The integral from 0 to infinity of (e^{-9x}) is 1/9