What is the integral of 0.2e^{-0.2x} ?

The integral of 0.2e^{-0.2x} is -e^{-0.2x}+C

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integral of 0.2e^{-0.2x}

\int 0.2 e^{- 0.2 x} d x

- e^{- 0.2 x} + C

Solution steps

\int 0.2 e^{- 0.2 x} d x

Take the constant out:

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

= 0.2 \cdot \int e^{- 0.2 x} d x

Apply u-substitution

: \int - \frac{1}{0.2} e^{u} d u

= 0.2 \cdot \int - \frac{1}{0.2} e^{u} d u

Take the constant out:

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

= 0.2 (- \frac{1}{0.2} \cdot \int e^{u} d u)

Simplify

= 0.2 (- 5 \cdot \int e^{u} d u)

Use the common integral:

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

= 0.2 (- 5 e^{u})

Substitute back

u = - 0.2 x

= 0.2 (- 5 e^{- 0.2 x})

Simplify

= - e^{- 0.2 x}

Add a constant to the solution

= - e^{- 0.2 x} + C

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What is the integral of 0.2e^{-0.2x} ?
The integral of 0.2e^{-0.2x} is -e^{-0.2x}+C