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

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

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

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

1 - \frac{1}{e}

Decimal

0.63212 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{- 1} - 2 e^{u} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Use the common integral:

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

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

Simplify

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

= [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 1 - \frac{1}{e}

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