What is the integral from 0 to infinity of 8e^{-4x} ?

The integral from 0 to infinity of 8e^{-4x} is 2

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integral from 0 to infinity of 8e^{-4x}

\int_{0}^{\infty} 8 e^{- 4 x} d x

2

Solution steps

\int_{0}^{\infty} 8 e^{- 4 x} d x

Compute the indefinite integral:

\int 8 e^{- 4 x} d x = - 2 e^{- 4 x} + C

Compute the boundaries:

\int_{0}^{\infty} 8 e^{- 4 x} d x = 0 - (- 2)

= 0 - (- 2)

Simplify

= 2

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What is the integral from 0 to infinity of 8e^{-4x} ?
The integral from 0 to infinity of 8e^{-4x} is 2