What is the integral from 2 to 6 of e^{-x/2} ?

The integral from 2 to 6 of e^{-x/2} is 2(1/e-1/(e^3))

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integral from 2 to 6 of e^{-x/2}

\int_{2}^{6} e^{- \frac{x}{2}} d x

2 (\frac{1}{e} - \frac{1}{e ^{3}})

Decimal

0.63618 \dots

Solution steps

\int_{2}^{6} e^{- \frac{x}{2}} d x

Apply u-substitution

= \int_{- 1}^{- 3} - 2 e^{u} d u

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

= - \int_{- 3}^{- 1} - 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_{- 3}^{- 1} e^{u} d u)

Use the common integral:

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

= - (- 2 [e^{u}]_{- 3}^{- 1})

Simplify

= 2 [e^{u}]_{- 3}^{- 1}

Compute the boundaries:

\frac{1}{e} - \frac{1}{e ^{3}}

= 2 (\frac{1}{e} - \frac{1}{e ^{3}})

What is the integral from 2 to 6 of e^{-x/2} ?
The integral from 2 to 6 of e^{-x/2} is 2(1/e-1/(e^3))