What is the integral of (4e^{-0.8x}) ?

The integral of (4e^{-0.8x}) is -5e^{-0.8x}+C

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integral of (4e^{-0.8x})

\int (4 e^{- 0.8 x}) d x

- 5 e^{- 0.8 x} + C

Solution steps

\int 4 e^{- 0.8 x} d x

Take the constant out:

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

= 4 \cdot \int e^{- 0.8 x} d x

Apply u-substitution

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

= 4 \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

= 4 (- \frac{1}{0.8} \cdot \int e^{u} d u)

Simplify

= 4 (- 1.25 \cdot \int e^{u} d u)

Use the common integral:

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

= 4 (- 1.25 e^{u})

Substitute back

u = - 0.8 x

= 4 (- 1.25 e^{- 0.8 x})

Simplify

= - 5 e^{- 0.8 x}

Add a constant to the solution

= - 5 e^{- 0.8 x} + C

\frac{d x}{d t} = e^{- k t}

x^{2}, x^{3}

\frac{\partial}{\partial y} ((x^{2} + y^{2})^{\frac{3}{2}})

\frac{\partial}{\partial x} (arctan (\frac{y}{x + a}))

f (x) = 8 x^{2} - x^{3}, at x = 1

What is the integral of (4e^{-0.8x}) ?
The integral of (4e^{-0.8x}) is -5e^{-0.8x}+C