What is the integral from 0 to 18 of 0.01e^{-0.01x} ?

The integral from 0 to 18 of 0.01e^{-0.01x} is 0.16472…

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integral from 0 to 18 of 0.01e^{-0.01x}

\int_{0}^{18} 0.01 e^{- 0.01 x} d x

0.16472 \dots

Solution steps

\int_{0}^{18} 0.01 e^{- 0.01 x} d x

Take the constant out:

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

= 0.01 \cdot \int_{0}^{18} e^{- 0.01 x} d x

Apply u-substitution

: \int_{0}^{- 0.18} - \frac{1}{0.01} e^{u} d u

= 0.01 \cdot \int_{0}^{- 0.18} - \frac{1}{0.01} e^{u} d u

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

= 0.01 (- \int_{- 0.18}^{0} - \frac{1}{0.01} e^{u} d u)

Take the constant out:

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

= 0.01 (- (- \frac{1}{0.01} \cdot \int_{- 0.18}^{0} e^{u} d u))

Simplify

= 0.01 (- (- 100 \cdot \int_{- 0.18}^{0} e^{u} d u))

Use the common integral:

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

= 0.01 (- (- 100 [e^{u}]_{- 0.18}^{0}))

Simplify

= [e^{u}]_{- 0.18}^{0}

Compute the boundaries:

0.16472 \dots

= 0.16472 \dots

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What is the integral from 0 to 18 of 0.01e^{-0.01x} ?
The integral from 0 to 18 of 0.01e^{-0.01x} is 0.16472…