What is the integral of-8e^{-0.1x} ?

The integral of-8e^{-0.1x} is 80e^{-0.1x}+C

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integral of-8e^{-0.1x}

\int - 8 e^{- 0.1 x} d x

80 e^{- 0.1 x} + C

Solution steps

\int - 8 e^{- 0.1 x} d x

Take the constant out:

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

= - 8 \cdot \int e^{- 0.1 x} d x

Apply u-substitution

: \int - \frac{1}{0.1} e^{u} d u

= - 8 \cdot \int - \frac{1}{0.1} e^{u} d u

Take the constant out:

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

= - 8 (- \frac{1}{0.1} \cdot \int e^{u} d u)

Simplify

= - 8 (- 10 \cdot \int e^{u} d u)

Use the common integral:

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

= - 8 (- 10 e^{u})

Substitute back

u = - 0.1 x

= - 8 (- 10 e^{- 0.1 x})

Simplify

= 80 e^{- 0.1 x}

Add a constant to the solution

= 80 e^{- 0.1 x} + C

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What is the integral of-8e^{-0.1x} ?
The integral of-8e^{-0.1x} is 80e^{-0.1x}+C