What is the integral from 5 to 15 of 0.02e^{-0.02x} ?

The integral from 5 to 15 of 0.02e^{-0.02x} is 0.16401…

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integral from 5 to 15 of 0.02e^{-0.02x}

\int_{5}^{15} 0.02 e^{- 0.02 x} d x

0.16401 \dots

Solution steps

\int_{5}^{15} 0.02 e^{- 0.02 x} d x

Take the constant out:

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

= 0.02 \cdot \int_{5}^{15} e^{- 0.02 x} d x

Apply u-substitution

: \int_{- 0.1}^{- 0.3} - \frac{1}{0.02} e^{u} d u

= 0.02 \cdot \int_{- 0.1}^{- 0.3} - \frac{1}{0.02} e^{u} d u

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

= 0.02 (- \int_{- 0.3}^{- 0.1} - \frac{1}{0.02} e^{u} d u)

Take the constant out:

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

= 0.02 (- (- \frac{1}{0.02} \cdot \int_{- 0.3}^{- 0.1} e^{u} d u))

Simplify

= 0.02 (- (- 50 \cdot \int_{- 0.3}^{- 0.1} e^{u} d u))

Use the common integral:

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

= 0.02 (- (- 50 [e^{u}]_{- 0.3}^{- 0.1}))

Simplify

= [e^{u}]_{- 0.3}^{- 0.1}

Compute the boundaries:

0.16401 \dots

= 0.16401 \dots

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What is the integral from 5 to 15 of 0.02e^{-0.02x} ?
The integral from 5 to 15 of 0.02e^{-0.02x} is 0.16401…