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

The integral from 0 to infinity of e^{-3t} is 1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{0}^{\infty} e^{- 3 t} d t

\frac{1}{3}

Solution steps

\int_{0}^{\infty} e^{- 3 t} d t

Compute the indefinite integral:

\int e^{- 3 t} d t = - \frac{1}{3} e^{- 3 t} + C

Compute the boundaries:

\int_{0}^{\infty} e^{- 3 t} d t = 0 - (- \frac{1}{3})

= 0 - (- \frac{1}{3})

Simplify

= \frac{1}{3}

\int_{0}^{2} \frac{1}{ln ( x )} d x

\int_{0}^{l n (3)} 6 e^{3 x} d x

\int_{e}^{e^{2}} x ln (x) d x

\int_{0}^{1} - 2 x + 2 d x

\int_{- 1}^{0} x^{2} (4 x^{3} + 5)^{3} d x

What is the integral from 0 to infinity of e^{-3t} ?
The integral from 0 to infinity of e^{-3t} is 1/3