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

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

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integral of e^{-0.01t}

\int e^{- 0.01 t} d t

- 100 e^{- 0.01 t} + C

Solution steps

\int e^{- 0.01 t} d t

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 t

= - 100 e^{- 0.01 t}

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 3 ∣})

\frac{\partial}{\partial y} (2 x - 3 yz)

\frac{d}{d r} (\frac{200 r ( r + 2 )}{( 2 + 3 r ) ^{2}})

y = 9 - 10 x, at x = - 1

\frac{d}{d x} (2 x y^{3})

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