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

The integral from θ to x of e^{-(x-θ)} is -e^{θ-x}+1

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integral from θ to x of e^{-(x-θ)}

\int_{θ}^{x} e^{- (x - θ)} d x

- e^{θ - x} + 1

Solution steps

\int_{θ}^{x} e^{- (x - θ)} d x

Simplify

= \int_{θ}^{x} e^{θ - x} d x

Apply u-substitution

: \int_{0}^{θ - x} - e^{u} d u

= \int_{0}^{θ - x} - e^{u} d u

Take the constant out:

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

= - \int_{0}^{θ - x} e^{u} d u

Use the common integral:

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

= - [e^{u}]_{0}^{θ - x}

Compute the boundaries:

e^{θ - x} - 1

= - (e^{θ - x} - 1)

Simplify

= - e^{θ - x} + 1

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