What is the integral from 2 to infinity of e^{-9p} ?

The integral from 2 to infinity of e^{-9p} is 1/(9e^{18)}

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integral from 2 to infinity of e^{-9p}

\int_{2}^{\infty} e^{- 9 p} d p

\frac{1}{9 e ^{18}}

Decimal Notation

1.69222 E - 9

Solution steps

\int_{2}^{\infty} e^{- 9 p} d p

Compute the indefinite integral:

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

Compute the boundaries:

\int_{2}^{\infty} e^{- 9 p} d p = 0 - (- \frac{1}{9 e ^{18}})

= 0 - (- \frac{1}{9 e ^{18}})

Simplify

= \frac{1}{9 e ^{18}}

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What is the integral from 2 to infinity of e^{-9p} ?
The integral from 2 to infinity of e^{-9p} is 1/(9e^{18)}