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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int - 4 e^{- \frac{t}{2}} t d t = - 4 (- 2 e^{- \frac{t}{2}} t - 4 e^{- \frac{t}{2}})

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

Simplify

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

Compute the boundaries:

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

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

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