What is the integral from 2 to infinity of e^{-6x} ?

The integral from 2 to infinity of e^{-6x} is 1/(6e^{12)}

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integral from 2 to infinity of e^{-6x}

\int_{2}^{\infty} e^{- 6 x} d x

\frac{1}{6 e ^{12}}

Decimal Notation

1.02404 E - 6

Solution steps

\int_{2}^{\infty} e^{- 6 x} d x

Compute the indefinite integral:

\int e^{- 6 x} d x = - \frac{1}{6} e^{- 6 x} + C

Compute the boundaries:

\int_{2}^{\infty} e^{- 6 x} d x = 0 - (- \frac{1}{6 e ^{12}})

= 0 - (- \frac{1}{6 e ^{12}})

Simplify

= \frac{1}{6 e ^{12}}

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What is the integral from 2 to infinity of e^{-6x} ?
The integral from 2 to infinity of e^{-6x} is 1/(6e^{12)}