What is the integral from 0 to t of xe^{t-x} ?

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

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integral from 0 to t of xe^{t-x}

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

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

Solution steps

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

Simplify

e^{t - x} : e^{t} e^{- x}

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

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^{- x} x d x

Apply Integration By Parts

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

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

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

= e^{t} [- e^{- x} x - e^{- x}]_{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 xe^{t-x} ?
The integral from 0 to t of xe^{t-x} is e^t(-te^{-t}-e^{-t}+1)