What is the integral from 0 to t of μe^{t-μ} ?

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

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integral from 0 to t of μe^{t-μ}

\int_{0}^{t} μ e^{t - μ} d μ

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

Solution steps

\int_{0}^{t} μ e^{t - μ} d μ

Simplify

e^{t - μ} : e^{t} e^{- μ}

= \int_{0}^{t} e^{t} e^{- μ} μ d μ

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^{- μ} μ d μ

Apply Integration By Parts

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

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

\int - e^{- μ} d μ = e^{- μ}

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