What is the integral from 0 to t of re^{t-r} ?

The integral from 0 to t of re^{t-r} is e^t(-te^{-t}-e^{-t}+1)

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integral from 0 to t of re^{t-r}

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

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

Solution steps

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

Simplify

e^{t - r} : e^{t} e^{- r}

= \int_{0}^{t} e^{t} e^{- r} r d r

Take the constant out:

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

= e^{t} \cdot \int_{0}^{t} e^{- r} r d r

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

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What is the integral from 0 to t of re^{t-r} ?
The integral from 0 to t of re^{t-r} is e^t(-te^{-t}-e^{-t}+1)