What is the integral of t^2e^{-t} ?

The integral of t^2e^{-t} is -e^{-t}t^2+2(-e^{-t}t-e^{-t})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of t^2e^{-t}

\int t^{2} e^{- t} d t

- e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t}) + C

Solution steps

\int t^{2} e^{- t} d t

Apply Integration By Parts

: - e^{- t} t^{2} - \int - 2 e^{- t} t d t

= - e^{- t} t^{2} - \int - 2 e^{- t} t d t

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

= - e^{- t} t^{2} - (- 2 (- e^{- t} t - e^{- t}))

Simplify

= - e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t})

Add a constant to the solution

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

x \to 2 lim (\frac{1}{x ^{2} - 1})

y = \frac{1}{8 x ^{2}}, y = x, y = \frac{x}{8}

\frac{d}{d x} (x sec (y))

\frac{1}{x - 4}

\int_{0}^{4} (x - (x - 2)) d x

What is the integral of t^2e^{-t} ?
The integral of t^2e^{-t} is -e^{-t}t^2+2(-e^{-t}t-e^{-t})+C