What is the integral of 20e^{-0.6t} ?

The integral of 20e^{-0.6t} is -33.33333…e^{-0.6t}+C

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integral of 20e^{-0.6t}

\int 20 e^{- 0.6 t} d t

- 33.33333 \dots e^{- 0.6 t} + C

Solution steps

\int 20 e^{- 0.6 t} d t

Take the constant out:

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

= 20 \cdot \int e^{- 0.6 t} d t

Apply u-substitution

: \int - \frac{1}{0.6} e^{u} d u

= 20 \cdot \int - \frac{1}{0.6} e^{u} d u

Take the constant out:

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

= 20 (- \frac{1}{0.6} \cdot \int e^{u} d u)

Simplify

= 20 (- 1.66666 \dots \cdot \int e^{u} d u)

Use the common integral:

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

= 20 (- 1.66666 \dots e^{u})

Substitute back

u = - 0.6 t

= 20 (- 1.66666 \dots e^{- 0.6 t})

Simplify

= - 33.33333 \dots e^{- 0.6 t}

Add a constant to the solution

= - 33.33333 \dots e^{- 0.6 t} + C

\int_{0}^{4} (2 x + 3) d x

\frac{d}{d x} (ln (\frac{x ^{2} + 1}{x + 1}))

f (x) = (x^{3} - 9 x)^{12}, at x = 3

\frac{9}{t ^{4}}

What is the integral of 20e^{-0.6t} ?
The integral of 20e^{-0.6t} is -33.33333…e^{-0.6t}+C