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

The integral from 0 to infinity of x^2e^{(-x)/2} is 16

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 x^2e^{(-x)/2}

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

16

Solution steps

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

Compute the indefinite integral:

\int x^{2} e^{\frac{- x}{2}} d x = - 2 e^{- \frac{x}{2}} x^{2} + 4 (- 2 e^{- \frac{x}{2}} x - 4 e^{- \frac{x}{2}}) + C

Compute the boundaries:

\int_{0}^{\infty} x^{2} e^{\frac{- x}{2}} d x = 0 - (- 16)

= 0 - (- 16)

Simplify

= 16

\int_{0}^{1} e^{- ik x} d x

\int_{0}^{0.75} 2 - 2 x d x

\int_{- 1}^{1} (x^{4} - x^{6}) d x

\int_{0}^{3} 5^{x} d x

\int_{1}^{3} x^{2} - 4 x d x

What is the integral from 0 to infinity of x^2e^{(-x)/2} ?
The integral from 0 to infinity of x^2e^{(-x)/2} is 16