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

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

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

2

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- 2)

Simplify

= 2

\int e^{- \frac{1}{2} x^{2}} d x

y^{^{''}} + 2 y + y = 2 e^{- t}

\frac{x}{x - 2}

\int_{2}^{5} x^{4} ln (2 x) d x

\int_{0}^{18} t^{2} + 2 t + 1 d t

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