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

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

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

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

1

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- 1)

Simplify

= 1

d er i v a t i v e \frac{1}{2 y}

a re a y = \frac{40}{x}, y = - 9 x + 39

\frac{d}{d x} (6 x - 2 x)

\int_{- 5}^{6} 0

\frac{d}{d x} (- 6 sin (x) - 6 x cos (x) + x^{2} sin (x))

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