What is the integral from 0 to x of 1/2 e^{-t/2} ?

The integral from 0 to x of 1/2 e^{-t/2} is -e^{-x/2}+1

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integral from 0 to x of 1/2 e^{-t/2}

\int_{0}^{x} \frac{1}{2} e^{- \frac{t}{2}} d t

- e^{- \frac{x}{2}} + 1

Solution steps

\int_{0}^{x} \frac{1}{2} e^{- \frac{t}{2}} d t

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{x} e^{- \frac{t}{2}} d t

Apply u-substitution

: \int_{0}^{- \frac{x}{2}} - 2 e^{u} d u

= \frac{1}{2} \cdot \int_{0}^{- \frac{x}{2}} - 2 e^{u} d u

Take the constant out:

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

= \frac{1}{2} (- 2 \cdot \int_{0}^{- \frac{x}{2}} e^{u} d u)

Use the common integral:

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

= \frac{1}{2} (- 2 [e^{u}]_{0}^{- \frac{x}{2}})

Simplify

\frac{1}{2} (- 2 [e^{u}]_{0}^{- \frac{x}{2}}) : - [e^{u}]_{0}^{- \frac{x}{2}}

= - [e^{u}]_{0}^{- \frac{x}{2}}

Compute the boundaries:

e^{- \frac{x}{2}} - 1

= - (e^{- \frac{x}{2}} - 1)

Simplify

= - e^{- \frac{x}{2}} + 1

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What is the integral from 0 to x of 1/2 e^{-t/2} ?
The integral from 0 to x of 1/2 e^{-t/2} is -e^{-x/2}+1