What is the integral from 0 to infinity of e^{-2t} ?

The integral from 0 to infinity of e^{-2t} is 1/2

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^{-2t}

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Compute the indefinite integral:

\int e^{- 2 t} d t = - \frac{1}{2} e^{- 2 t} + C

Compute the boundaries:

\int_{0}^{\infty} e^{- 2 t} d t = 0 - (- \frac{1}{2})

= 0 - (- \frac{1}{2})

Simplify

= \frac{1}{2}

\int (x + 2) (x^{2} + 4 x + 2)^{10} d x

\int - 64 d x

n = 1 \sum \infty (- 0.2)^{n - 1}

\frac{d}{d x} (x^{2} 2 x - 7)

\int \frac{e ^{8 x}}{e ^{8 x} + 7} d x

What is the integral from 0 to infinity of e^{-2t} ?
The integral from 0 to infinity of e^{-2t} is 1/2