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

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

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

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

- e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t}) + 2

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

- e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t}) + 2

= - e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t}) + 2

\int (t - 3) cos (t - 3) d t

\frac{\partial}{\partial x} (x e^{x y})

\int_{- 6}^{6} \frac{1}{36 - x ^{2}} d x

\int \frac{1}{( 3 x + 1 )} d x

x \to 3 + lim (\frac{x + 4}{x - 3})

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