What is the integral of t^2e^{-22t} ?

The integral of t^2e^{-22t} is -1/5324 e^{-22t}(242t^2+22t+1)+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^{-22t}

\int t^{2} e^{- 22 t} d t

- \frac{1}{5324} e^{- 22 t} (242 t^{2} + 22 t + 1) + C

Solution steps

\int t^{2} e^{- 22 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{10648} d u

= \int - \frac{e ^{u} u ^{2}}{10648} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{10648} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= - \frac{1}{10648} (u^{2} e^{u} - \int 2 u e^{u} d u)

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

= - \frac{1}{10648} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 22 t

= - \frac{1}{10648} ((- 22 t)^{2} e^{- 22 t} - 2 (e^{- 22 t} (- 22 t) - e^{- 22 t}))

Simplify

- \frac{1}{10648} ((- 22 t)^{2} e^{- 22 t} - 2 (e^{- 22 t} (- 22 t) - e^{- 22 t})) : - \frac{1}{5324} e^{- 22 t} (242 t^{2} + 22 t + 1)

= - \frac{1}{5324} e^{- 22 t} (242 t^{2} + 22 t + 1)

Add a constant to the solution

= - \frac{1}{5324} e^{- 22 t} (242 t^{2} + 22 t + 1) + C

\int 1 - 7 x^{2} d x

n = 2 \sum \infty \frac{5}{2 ^{n}}

t e^{- 5 t}

5 t ln (t)

\int_{0}^{b} \frac{1}{16 - x ^{2}} d x

What is the integral of t^2e^{-22t} ?
The integral of t^2e^{-22t} is -1/5324 e^{-22t}(242t^2+22t+1)+C