What is the integral from 0 to t of ue^{-u} ?

The integral from 0 to t of ue^{-u} is -te^{-t}-e^{-t}+1

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integral from 0 to t of ue^{-u}

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

- t e^{- t} - e^{- t} + 1

Solution steps

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

Apply Integration By Parts

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

\int - e^{- u} d u = e^{- u}

= [- e^{- u} u - e^{- u}]_{0}^{t}

Compute the boundaries:

- t e^{- t} - e^{- t} + 1

= - t e^{- t} - e^{- t} + 1

What is the integral from 0 to t of ue^{-u} ?
The integral from 0 to t of ue^{-u} is -te^{-t}-e^{-t}+1