What is the integral from 0 to infinity of 6e^{-3x} ?

The integral from 0 to infinity of 6e^{-3x} is 2

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integral from 0 to infinity of 6e^{-3x}

\int_{0}^{\infty} 6 e^{- 3 x} d x

2

Solution steps

\int_{0}^{\infty} 6 e^{- 3 x} d x

Compute the indefinite integral:

\int 6 e^{- 3 x} d x = - 2 e^{- 3 x} + C

Compute the boundaries:

\int_{0}^{\infty} 6 e^{- 3 x} d x = 0 - (- 2)

= 0 - (- 2)

Simplify

= 2

\frac{d}{d x} (\frac{x - 19}{x - 19})

f (x) = x^{4} - 13 x^{2} + 36, at x = 1

(25 t) ((\frac{t}{25})^{2})

\frac{d}{d x} (x^{2} + \frac{1}{2 x} - 3 x)

n = 0 \sum \infty \frac{1}{2 ( 2 ) ^{n}}

What is the integral from 0 to infinity of 6e^{-3x} ?
The integral from 0 to infinity of 6e^{-3x} is 2