What is the integral of 0.02e^{-0.02x} ?

The integral of 0.02e^{-0.02x} is -e^{-0.02x}+C

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integral of 0.02e^{-0.02x}

\int 0.02 e^{- 0.02 x} d x

- e^{- 0.02 x} + C

Solution steps

\int 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 e^{- 0.02 x} d x

Apply u-substitution

: \int - \frac{1}{0.02} e^{u} d u

= 0.02 \cdot \int - \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 e^{u} d u)

Simplify

= 0.02 (- 50 \cdot \int e^{u} d u)

Use the common integral:

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

= 0.02 (- 50 e^{u})

Substitute back

u = - 0.02 x

= 0.02 (- 50 e^{- 0.02 x})

Simplify

= - e^{- 0.02 x}

Add a constant to the solution

= - e^{- 0.02 x} + C

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What is the integral of 0.02e^{-0.02x} ?
The integral of 0.02e^{-0.02x} is -e^{-0.02x}+C