What is the integral from 2 to infinity of e^{-8p} ?

The integral from 2 to infinity of e^{-8p} is 1/(8e^{16)}

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integral from 2 to infinity of e^{-8p}

\int_{2}^{\infty} e^{- 8 p} d p

\frac{1}{8 e ^{16}}

Decimal

1.40669 E - 8

Solution steps

\int_{2}^{\infty} e^{- 8 p} d p

Compute the indefinite integral:

\int e^{- 8 p} d p = - \frac{1}{8} e^{- 8 p} + C

Compute the boundaries:

\int_{2}^{\infty} e^{- 8 p} d p = 0 - (- \frac{1}{8 e ^{16}})

= 0 - (- \frac{1}{8 e ^{16}})

Simplify

= \frac{1}{8 e ^{16}}

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What is the integral from 2 to infinity of e^{-8p} ?
The integral from 2 to infinity of e^{-8p} is 1/(8e^{16)}