What is the integral of e^{-0.01x} ?

The integral of e^{-0.01x} is -100e^{-0.01x}+C

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integral of e^{-0.01x}

\int e^{- 0.01 x} d x

- 100 e^{- 0.01 x} + C

Solution steps

\int e^{- 0.01 x} d x

Apply u-substitution

: \int - \frac{1}{0.01} e^{u} d u

= \int - \frac{1}{0.01} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.01} \cdot \int e^{u} d u

Simplify

= - 100 \cdot \int e^{u} d u

Use the common integral:

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

= - 100 e^{u}

Substitute back

u = - 0.01 x

= - 100 e^{- 0.01 x}

Add a constant to the solution

= - 100 e^{- 0.01 x} + C

10 sin^{2} (x) + 10 cos^{2} (x)

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\frac{1}{2 x} y^{^{'}} = \frac{x}{3 y}

\frac{d y}{d x} = (y^{2} - 9)

What is the integral of e^{-0.01x} ?
The integral of e^{-0.01x} is -100e^{-0.01x}+C