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

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

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

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

8

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- 8)

Simplify

= 8

\int_{0}^{5} \frac{4}{x ^{2}} d x

\frac{d}{d x} (e^{- x} + e^{x})

n = 4 \sum \infty \frac{1}{n ^{2} + n}

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

x \to 9 + lim (\frac{x - 9}{∣ x - 9 ∣})

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