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

The integral of t^2e^{-t/2} is -2e^{-t/2}t^2+4(-2e^{-t/2}t-4e^{-t/2})+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/2}

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

- 2 e^{- \frac{t}{2}} t^{2} + 4 (- 2 e^{- \frac{t}{2}} t - 4 e^{- \frac{t}{2}}) + C

Solution steps

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

Apply Integration By Parts

: - 2 e^{- \frac{t}{2}} t^{2} - \int - 4 e^{- \frac{t}{2}} t d t

= - 2 e^{- \frac{t}{2}} t^{2} - \int - 4 e^{- \frac{t}{2}} t d t

\int - 4 e^{- \frac{t}{2}} t d t = - 4 (- 2 e^{- \frac{t}{2}} t - 4 e^{- \frac{t}{2}})

= - 2 e^{- \frac{t}{2}} t^{2} - (- 4 (- 2 e^{- \frac{t}{2}} t - 4 e^{- \frac{t}{2}}))

Simplify

= - 2 e^{- \frac{t}{2}} t^{2} + 4 (- 2 e^{- \frac{t}{2}} t - 4 e^{- \frac{t}{2}})

Add a constant to the solution

= - 2 e^{- \frac{t}{2}} t^{2} + 4 (- 2 e^{- \frac{t}{2}} t - 4 e^{- \frac{t}{2}}) + C

\frac{d}{d x} (7 x^{2} - 5 x + 9)

\int \frac{1}{4 x ^{4}} d x

\frac{d}{d x} (- 5 x^{2} - 4 cos (x))

\int 2 x e^{- 8 x} d x

n = 0 \sum \infty 300 (- 1.41)^{n}

What is the integral of t^2e^{-t/2} ?
The integral of t^2e^{-t/2} is -2e^{-t/2}t^2+4(-2e^{-t/2}t-4e^{-t/2})+C