What is the integral from 1 to infinity of 8e^{-8x} ?

The integral from 1 to infinity of 8e^{-8x} is 1/(e^8)

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integral from 1 to infinity of 8e^{-8x}

\int_{1}^{\infty} 8 e^{- 8 x} d x

\frac{1}{e ^{8}}

Decimal

0.00033 \dots

Solution steps

\int_{1}^{\infty} 8 e^{- 8 x} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} 8 e^{- 8 x} d x = 0 - (- \frac{1}{e ^{8}})

= 0 - (- \frac{1}{e ^{8}})

Simplify

= \frac{1}{e ^{8}}

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What is the integral from 1 to infinity of 8e^{-8x} ?
The integral from 1 to infinity of 8e^{-8x} is 1/(e^8)