What is the integral of e^{-0.05x} ?

The integral of e^{-0.05x} is -20e^{-0.05x}+C

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integral of e^{-0.05x}

\int e^{- 0.05 x} d x

- 20 e^{- 0.05 x} + C

Solution steps

\int e^{- 0.05 x} d x

Apply u-substitution

: \int - \frac{1}{0.05} e^{u} d u

= \int - \frac{1}{0.05} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.05} \cdot \int e^{u} d u

Simplify

= - 20 \cdot \int e^{u} d u

Use the common integral:

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

= - 20 e^{u}

Substitute back

u = - 0.05 x

= - 20 e^{- 0.05 x}

Add a constant to the solution

= - 20 e^{- 0.05 x} + C

y^{^{'}} + 3 = y + 3 x

\frac{1}{s ^{2} + s - 20}

y = x x, (16, 64)

\int 36 - x^{2} d x

\frac{d ^{2}}{d x ^{2}} (\frac{6}{x})

What is the integral of e^{-0.05x} ?
The integral of e^{-0.05x} is -20e^{-0.05x}+C