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

The integral of 2t^2e^{-t} is 2(-e^{-t}t^2+2(-e^{-t}t-e^{-t}))+C

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integral of 2t^2e^{-t}

\int 2 t^{2} e^{- t} d t

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

Solution steps

\int 2 t^{2} e^{- t} d t

Take the constant out:

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

= 2 \cdot \int t^{2} e^{- t} d t

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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