What is the integral from 6 to infinity of e^{-3x} ?

The integral from 6 to infinity of e^{-3x} is 1/(3e^{18)}

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integral from 6 to infinity of e^{-3x}

\int_{6}^{\infty} e^{- 3 x} d x

\frac{1}{3 e ^{18}}

Decimal

5.07666 E - 9

Solution steps

\int_{6}^{\infty} e^{- 3 x} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{6}^{\infty} e^{- 3 x} d x = 0 - (- \frac{1}{3 e ^{18}})

= 0 - (- \frac{1}{3 e ^{18}})

Simplify

= \frac{1}{3 e ^{18}}

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What is the integral from 6 to infinity of e^{-3x} ?
The integral from 6 to infinity of e^{-3x} is 1/(3e^{18)}