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

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

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

2

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- 2)

Simplify

= 2

\int_{0}^{4} x cos (\frac{nπ x}{4}) d x

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

\int_{t}^{3 t} (\frac{sin ( x )}{3 x}) d x

\int_{\frac{π}{3}}^{\frac{π}{2}} csc^{2} (\frac{x}{2}) d x

\int_{0}^{1} \frac{x ( 1 + 3 y ^{2} )}{4} d y

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