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

The integral from 0 to t of t^2e^{-4t} is -(8e^{-4t}t^2+e^{-4t}(4t+1)-1)/(32)

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

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

- \frac{8 e ^{- 4 t} t ^{2} + e ^{- 4 t} ( 4 t + 1 ) - 1}{32}

Solution steps

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

Apply u-substitution

: \int_{0}^{- 4 t} - \frac{e ^{u} u ^{2}}{64} d u

= \int_{0}^{- 4 t} - \frac{e ^{u} u ^{2}}{64} d u

Take the constant out:

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

= - \frac{1}{64} \cdot \int_{0}^{- 4 t} e^{u} u^{2} d u

Apply Integration By Parts

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

= - \frac{1}{64} [u^{2} e^{u} - \int 2 u e^{u} d u]_{0}^{- 4 t}

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

= - \frac{1}{64} [u^{2} e^{u} - 2 (e^{u} u - e^{u})]_{0}^{- 4 t}

Compute the boundaries:

16 e^{- 4 t} t^{2} - 2 (- 4 e^{- 4 t} t - e^{- 4 t}) - 2

= - \frac{1}{64} (16 e^{- 4 t} t^{2} - 2 (- 4 e^{- 4 t} t - e^{- 4 t}) - 2)

Simplify

= - \frac{8 e ^{- 4 t} t ^{2} + e ^{- 4 t} ( 4 t + 1 ) - 1}{32}

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What is the integral from 0 to t of t^2e^{-4t} ?
The integral from 0 to t of t^2e^{-4t} is -(8e^{-4t}t^2+e^{-4t}(4t+1)-1)/(32)