What is the integral from-1 to 18 of e^{-x/2} ?

The integral from-1 to 18 of e^{-x/2} is 2(sqrt(e)-1/(e^9))

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integral from-1 to 18 of e^{-x/2}

\int_{- 1}^{18} e^{- \frac{x}{2}} d x

2 (e - \frac{1}{e ^{9}})

Decimal

3.29719 \dots

Solution steps

\int_{- 1}^{18} e^{- \frac{x}{2}} d x

Apply u-substitution

: \int_{\frac{1}{2}}^{- 9} - 2 e^{u} d u

= \int_{\frac{1}{2}}^{- 9} - 2 e^{u} d u

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

= - \int_{- 9}^{\frac{1}{2}} - 2 e^{u} d u

Take the constant out:

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

= - (- 2 \cdot \int_{- 9}^{\frac{1}{2}} e^{u} d u)

Use the common integral:

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

= - (- 2 [e^{u}]_{- 9}^{\frac{1}{2}})

Simplify

= 2 [e^{u}]_{- 9}^{\frac{1}{2}}

Compute the boundaries:

e - \frac{1}{e ^{9}}

= 2 (e - \frac{1}{e ^{9}})

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What is the integral from-1 to 18 of e^{-x/2} ?
The integral from-1 to 18 of e^{-x/2} is 2(sqrt(e)-1/(e^9))