What is the integral from 0 to t of e^{-(t-x)} ?

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

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integral from 0 to t of e^{-(t-x)}

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

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

Solution steps

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

Simplify

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

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{t} - 1

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

What is the integral from 0 to t of e^{-(t-x)} ?
The integral from 0 to t of e^{-(t-x)} is e^{-t}(e^t-1)