What is the integral from 1 to infinity of e^{-3n} ?

The integral from 1 to infinity of e^{-3n} is 1/(3e^3)

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integral from 1 to infinity of e^{-3n}

\int_{1}^{\infty} e^{- 3 n} d n

\frac{1}{3 e ^{3}}

Decimal Notation

0.01659 \dots

Solution steps

\int_{1}^{\infty} e^{- 3 n} d n

Compute the indefinite integral:

\int e^{- 3 n} d n = - \frac{1}{3} e^{- 3 n} + C

Compute the boundaries:

\int_{1}^{\infty} e^{- 3 n} d n = 0 - (- \frac{1}{3 e ^{3}})

= 0 - (- \frac{1}{3 e ^{3}})

Simplify

= \frac{1}{3 e ^{3}}

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What is the integral from 1 to infinity of e^{-3n} ?
The integral from 1 to infinity of e^{-3n} is 1/(3e^3)