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

The integral from 0 to infinity of 2e^{-x} is 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 2e^{-x}

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

2

Solution steps

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

Compute the indefinite integral:

\int 2 e^{- x} d x = - 2 e^{- x} + C

Compute the boundaries:

\int_{0}^{\infty} 2 e^{- x} d x = 0 - (- 2)

= 0 - (- 2)

Simplify

= 2

y^{^{'}} = \frac{y}{2} + 4 e^{\frac{t}{2}}

\frac{d}{d x} (\frac{x}{( x ^{2} - 1 )})

\frac{\partial}{\partial y} (ln (\frac{x}{x ^{2} + y ^{2}}))

\int t^{3} 1 + t^{2} d t

\int x 4^{x} d x

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