What is the integral from 1 to infinity of e^{-9x} ?

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

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integral from 1 to infinity of e^{-9x}

\int_{1}^{\infty} e^{- 9 x} d x

\frac{1}{9 e ^{9}}

Decimal Notation

0.00001 \dots

Solution steps

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

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

Simplify

= \frac{1}{9 e ^{9}}

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