What is the integral of 20e^{-0.8t} ?

The integral of 20e^{-0.8t} is -25e^{-0.8t}+C

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integral of 20e^{-0.8t}

\int 20 e^{- 0.8 t} d t

- 25 e^{- 0.8 t} + C

Solution steps

\int 20 e^{- 0.8 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^{- 0.8 t} d t

Apply u-substitution

: \int - \frac{1}{0.8} e^{u} d u

= 20 \cdot \int - \frac{1}{0.8} 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}{0.8} \cdot \int e^{u} d u)

Simplify

= 20 (- 1.25 \cdot \int e^{u} d u)

Use the common integral:

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

= 20 (- 1.25 e^{u})

Substitute back

u = - 0.8 t

= 20 (- 1.25 e^{- 0.8 t})

Simplify

= - 25 e^{- 0.8 t}

Add a constant to the solution

= - 25 e^{- 0.8 t} + C

\frac{\partial}{\partial x} (y + 2 - \frac{2}{x})

\frac{d}{d x} (3 sin (x) + cos (x))

x \to 8 lim (2 - 6 x^{2} + x^{3})

\int \frac{4 x ^{3} - 3 x}{x} d x

y^{2} y^{^{'}} = y^{3} - 3 x

What is the integral of 20e^{-0.8t} ?
The integral of 20e^{-0.8t} is -25e^{-0.8t}+C