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

The integral from 0 to infinity of e^{-bt} is 1/b

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integral from 0 to infinity of e^{-bt}

\int_{0}^{\infty} e^{- b t} d t

\frac{1}{b}

Solution steps

\int_{0}^{\infty} e^{- b t} d t

Compute the indefinite integral:

\int e^{- b t} d t = - \frac{1}{b} e^{- b t} + C

Compute the boundaries:

\int_{0}^{\infty} e^{- b t} d t = 0 - (- \frac{1}{b})

= 0 - (- \frac{1}{b})

Simplify

= \frac{1}{b}

\int \frac{x ^{2}}{4} - 1 d x

\frac{d}{d x} (2 e^{- x} x - e^{- x} x^{2})

\frac{d}{d x} ((4 x - x^{2})^{3})

y^{^{'}} + 8 y = 4, y (0) = 3

\int_{0}^{2} t e^{t^{2}} d t

What is the integral from 0 to infinity of e^{-bt} ?
The integral from 0 to infinity of e^{-bt} is 1/b