What is the integral from 0 to t of t^2e^{-3t} ?

The integral from 0 to t of t^2e^{-3t} is -1/27 (9e^{-3t}t^2-2(-3e^{-3t}t-e^{-3t})-2)

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integral from 0 to t of t^2e^{-3t}

\int_{0}^{t} t^{2} e^{- 3 t} d t

- \frac{1}{27} (9 e^{- 3 t} t^{2} - 2 (- 3 e^{- 3 t} t - e^{- 3 t}) - 2)

Solution steps

\int_{0}^{t} t^{2} e^{- 3 t} d t

Apply u-substitution

: \int_{0}^{- 3 t} - \frac{e ^{u} u ^{2}}{27} d u

= \int_{0}^{- 3 t} - \frac{e ^{u} u ^{2}}{27} d u

Take the constant out:

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

= - \frac{1}{27} \cdot \int_{0}^{- 3 t} e^{u} u^{2} d u

Apply Integration By Parts

: [u^{2} e^{u} - \int 2 u e^{u} d u]_{0}^{- 3 t}

= - \frac{1}{27} [u^{2} e^{u} - \int 2 u e^{u} d u]_{0}^{- 3 t}

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

= - \frac{1}{27} [u^{2} e^{u} - 2 (e^{u} u - e^{u})]_{0}^{- 3 t}

Compute the boundaries:

9 e^{- 3 t} t^{2} - 2 (- 3 e^{- 3 t} t - e^{- 3 t}) - 2

= - \frac{1}{27} (9 e^{- 3 t} t^{2} - 2 (- 3 e^{- 3 t} t - e^{- 3 t}) - 2)

\frac{\partial}{\partial x} (2 (y - 1) (x + 1))

\int_{- 1}^{7} 3 x + 7 - x^{2} + 3 x d x

\frac{d}{d x} (\frac{x ^{2} + 3 x}{x - 1})

\frac{d N}{d t} + N = Nt e^{t + 2}

What is the integral from 0 to t of t^2e^{-3t} ?
The integral from 0 to t of t^2e^{-3t} is -1/27 (9e^{-3t}t^2-2(-3e^{-3t}t-e^{-3t})-2)