What is the integral from 1 to infinity of 12e^{-6x} ?

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

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integral from 1 to infinity of 12e^{-6x}

\int_{1}^{\infty} 12 e^{- 6 x} d x

\frac{2}{e ^{6}}

Decimal Notation

0.00495 \dots

Solution steps

\int_{1}^{\infty} 12 e^{- 6 x} d x

Compute the indefinite integral:

\int 12 e^{- 6 x} d x = - 2 e^{- 6 x} + C

Compute the boundaries:

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

= 0 - (- \frac{2}{e ^{6}})

Simplify

= \frac{2}{e ^{6}}

\frac{10 s ^{2}}{s ^{2} + 9}

x \to 1 lim (x^{2} + 5)

y^{^{'}} = x + 4 y - 2

\int (x - 3)^{- 2} d x

x \to 0 lim (\frac{2 ^{x} - 3 ^{x}}{x})

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