What is the integral from 0 to infinity of e^{-5t} ?

The integral from 0 to infinity of e^{-5t} is 1/5

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integral from 0 to infinity of e^{-5t}

\int_{0}^{\infty} e^{- 5 t} d t

\frac{1}{5}

Decimal Notation

0.2

Solution steps

\int_{0}^{\infty} e^{- 5 t} d t

Compute the indefinite integral:

\int e^{- 5 t} d t = - \frac{1}{5} e^{- 5 t} + C

Compute the boundaries:

\int_{0}^{\infty} e^{- 5 t} d t = 0 - (- \frac{1}{5})

= 0 - (- \frac{1}{5})

Simplify

= \frac{1}{5}

f (x) = x (1 - x^{2})

y = (\frac{x ^{2} + 3}{x ^{2} - 3})^{8}

f (x) = \frac{7}{1 - x}

x^{2} x^{3} + 1

x \to 8 lim (x)

What is the integral from 0 to infinity of e^{-5t} ?
The integral from 0 to infinity of e^{-5t} is 1/5