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

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

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integral from 0 to x of 1/θ e^{(-x)/θ}

\int_{0}^{x} \frac{1}{θ} e^{\frac{- x}{θ}} d x

- e^{- \frac{x}{θ}} + 1

Solution steps

\int_{0}^{x} \frac{1}{θ} e^{\frac{- x}{θ}} d x

Take the constant out:

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

= \frac{1}{θ} \cdot \int_{0}^{x} e^{\frac{- x}{θ}} d x

Simplify

= \frac{1}{θ} \cdot \int_{0}^{x} e^{- \frac{x}{θ}} d x

Apply u-substitution

: \int_{0}^{- \frac{x}{θ}} - θ e^{u} d u

= \frac{1}{θ} \cdot \int_{0}^{- \frac{x}{θ}} - θ e^{u} d u

Take the constant out:

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

= \frac{1}{θ} (- θ \cdot \int_{0}^{- \frac{x}{θ}} e^{u} d u)

Use the common integral:

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

= \frac{1}{θ} (- θ [e^{u}]_{0}^{- \frac{x}{θ}})

Simplify

\frac{1}{θ} (- θ [e^{u}]_{0}^{- \frac{x}{θ}}) : - [e^{u}]_{0}^{- \frac{x}{θ}}

= - [e^{u}]_{0}^{- \frac{x}{θ}}

Compute the boundaries:

e^{- \frac{x}{θ}} - 1

= - (e^{- \frac{x}{θ}} - 1)

Simplify

= - e^{- \frac{x}{θ}} + 1

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