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

The integral from 0 to infinity of 8e^{-8x} is 1

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

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

1

Solution steps

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

= 0 - (- 1)

Simplify

= 1

y \to 1 lim (\frac{y - 1}{y ^{6} - 1})

ln (2 x + 3)

\frac{d}{d x} (\frac{8}{sin ( x )} + \frac{1}{cot ( x )})

\int_{0}^{\frac{π}{2}} cos^{2} (2 x) d x

sin^{4} (\frac{x}{4}) + cos^{4} (\frac{x}{4})

What is the integral from 0 to infinity of 8e^{-8x} ?
The integral from 0 to infinity of 8e^{-8x} is 1