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

The integral from 0 to t of t^2e^{-t/3} is -3e^{-t/3}t^2+6(-3e^{-t/3}t-9e^{-t/3})+54

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

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

- 3 e^{- \frac{t}{3}} t^{2} + 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}}) + 54

Solution steps

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

Apply Integration By Parts

: [- 3 e^{- \frac{t}{3}} t^{2} - \int - 6 e^{- \frac{t}{3}} t d t]_{0}^{t}

= [- 3 e^{- \frac{t}{3}} t^{2} - \int - 6 e^{- \frac{t}{3}} t d t]_{0}^{t}

\int - 6 e^{- \frac{t}{3}} t d t = - 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}})

= [- 3 e^{- \frac{t}{3}} t^{2} - (- 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}}))]_{0}^{t}

Simplify

= [- 3 e^{- \frac{t}{3}} t^{2} + 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}})]_{0}^{t}

Compute the boundaries:

- 3 e^{- \frac{t}{3}} t^{2} + 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}}) + 54

= - 3 e^{- \frac{t}{3}} t^{2} + 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}}) + 54

y = \frac{5}{x}, (5, 1)

\int_{1}^{8} (\frac{2}{x}) d x

f (x) = x^{3}, at x = 3

\int - 2 x e^{- \frac{1}{2} x} d x

\frac{\partial}{\partial x} (cos (x^{2}))

What is the integral from 0 to t of t^2e^{-t/3} ?
The integral from 0 to t of t^2e^{-t/3} is -3e^{-t/3}t^2+6(-3e^{-t/3}t-9e^{-t/3})+54