What is the integral of e^{-0.08x} ?

The integral of e^{-0.08x} is -12.5e^{-0.08x}+C

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integral of e^{-0.08x}

\int e^{- 0.08 x} d x

- 12.5 e^{- 0.08 x} + C

Solution steps

\int e^{- 0.08 x} d x

Apply u-substitution

: \int - \frac{1}{0.08} e^{u} d u

= \int - \frac{1}{0.08} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.08} \cdot \int e^{u} d u

Simplify

= - 12.5 \cdot \int e^{u} d u

Use the common integral:

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

= - 12.5 e^{u}

Substitute back

u = - 0.08 x

= - 12.5 e^{- 0.08 x}

Add a constant to the solution

= - 12.5 e^{- 0.08 x} + C

\frac{d}{d x} (tan^{2} (x) cos (x))

n = 0 \sum \infty 6 (\frac{3}{4})^{n}

\int (ln (11 x))^{2} d x

3 x + 7

- 2.25 cos (1.5 x)

What is the integral of e^{-0.08x} ?
The integral of e^{-0.08x} is -12.5e^{-0.08x}+C