What is the integral of t^2e^{-12t} ?

The integral of t^2e^{-12t} is -1/864 e^{-12t}(72t^2+12t+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^{-12t}

\int t^{2} e^{- 12 t} d t

- \frac{1}{864} e^{- 12 t} (72 t^{2} + 12 t + 1) + C

Solution steps

\int t^{2} e^{- 12 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{1728} d u

= \int - \frac{e ^{u} u ^{2}}{1728} d u

Take the constant out:

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

= - \frac{1}{1728} \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}{1728} (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}{1728} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 12 t

= - \frac{1}{1728} ((- 12 t)^{2} e^{- 12 t} - 2 (e^{- 12 t} (- 12 t) - e^{- 12 t}))

Simplify

- \frac{1}{1728} ((- 12 t)^{2} e^{- 12 t} - 2 (e^{- 12 t} (- 12 t) - e^{- 12 t})) : - \frac{1}{864} e^{- 12 t} (72 t^{2} + 12 t + 1)

= - \frac{1}{864} e^{- 12 t} (72 t^{2} + 12 t + 1)

Add a constant to the solution

= - \frac{1}{864} e^{- 12 t} (72 t^{2} + 12 t + 1) + C

\int \frac{x ^{2} + x - 6}{x ^{2} - 4 x + 3} d x

\frac{\partial}{\partial x} (\frac{4 x + y}{x - y})

x \to - \infty lim (3^{4 x})

\int_{0}^{0.5} 5 arccos (x) d x

y = \frac{1}{x ^{2}} + x, x = 2, x = 4

What is the integral of t^2e^{-12t} ?
The integral of t^2e^{-12t} is -1/864 e^{-12t}(72t^2+12t+1)+C