What is the integral from 0 to r of 0.5e^{-0.5x} ?

The integral from 0 to r of 0.5e^{-0.5x} is -e^{-0.5r}+1

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integral from 0 to r of 0.5e^{-0.5x}

\int_{0}^{r} 0.5 e^{- 0.5 x} d x

- e^{- 0.5 r} + 1

Solution steps

\int_{0}^{r} 0.5 e^{- 0.5 x} d x

Take the constant out:

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

= 0.5 \cdot \int_{0}^{r} e^{- 0.5 x} d x

Apply u-substitution

: \int_{0}^{- 0.5 r} - \frac{1}{0.5} e^{u} d u

= 0.5 \cdot \int_{0}^{- 0.5 r} - \frac{1}{0.5} e^{u} d u

Take the constant out:

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

= 0.5 (- \frac{1}{0.5} \cdot \int_{0}^{- 0.5 r} e^{u} d u)

Simplify

= 0.5 (- 2 \cdot \int_{0}^{- 0.5 r} e^{u} d u)

Use the common integral:

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

= 0.5 (- 2 [e^{u}]_{0}^{- 0.5 r})

Simplify

= - [e^{u}]_{0}^{- 0.5 r}

Compute the boundaries:

e^{- 0.5 r} - 1

= - (e^{- 0.5 r} - 1)

Simplify

= - e^{- 0.5 r} + 1

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What is the integral from 0 to r of 0.5e^{-0.5x} ?
The integral from 0 to r of 0.5e^{-0.5x} is -e^{-0.5r}+1