What is the integral from 0 to t of t^2e^{-t/4} ?

The integral from 0 to t of t^2e^{-t/4} is -4e^{-t/4}t^2+8(-4e^{-t/4}t-16e^{-t/4})+128

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t of t^2e^{-t/4}

\int_{0}^{t} t^{2} e^{- \frac{t}{4}} d t

- 4 e^{- \frac{t}{4}} t^{2} + 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}}) + 128

Solution steps

\int_{0}^{t} t^{2} e^{- \frac{t}{4}} d t

Apply Integration By Parts

: [- 4 e^{- \frac{t}{4}} t^{2} - \int - 8 e^{- \frac{t}{4}} t d t]_{0}^{t}

= [- 4 e^{- \frac{t}{4}} t^{2} - \int - 8 e^{- \frac{t}{4}} t d t]_{0}^{t}

\int - 8 e^{- \frac{t}{4}} t d t = - 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}})

= [- 4 e^{- \frac{t}{4}} t^{2} - (- 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}}))]_{0}^{t}

Simplify

= [- 4 e^{- \frac{t}{4}} t^{2} + 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}})]_{0}^{t}

Compute the boundaries:

- 4 e^{- \frac{t}{4}} t^{2} + 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}}) + 128

= - 4 e^{- \frac{t}{4}} t^{2} + 8 (- 4 e^{- \frac{t}{4}} t - 16 e^{- \frac{t}{4}}) + 128

\int (x + 6) (3 x - 4) d x

\int_{- 2}^{0} 3 x 4 - x^{2} d x

\frac{\partial}{\partial x} (ln (2 + x^{2} + y^{2}))

\frac{\partial}{\partial y} (ln (y - 3 x))

\int - 24 t e^{- 3 t} d t

What is the integral from 0 to t of t^2e^{-t/4} ?
The integral from 0 to t of t^2e^{-t/4} is -4e^{-t/4}t^2+8(-4e^{-t/4}t-16e^{-t/4})+128