What is the integral of (6e^{-0.2x}) ?

The integral of (6e^{-0.2x}) is -30e^{-0.2x}+C

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integral of (6e^{-0.2x})

\int (6 e^{- 0.2 x}) d x

- 30 e^{- 0.2 x} + C

Solution steps

\int 6 e^{- 0.2 x} d x

Take the constant out:

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

= 6 \cdot \int e^{- 0.2 x} d x

Apply u-substitution

: \int - \frac{1}{0.2} e^{u} d u

= 6 \cdot \int - \frac{1}{0.2} e^{u} d u

Take the constant out:

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

= 6 (- \frac{1}{0.2} \cdot \int e^{u} d u)

Simplify

= 6 (- 5 \cdot \int e^{u} d u)

Use the common integral:

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

= 6 (- 5 e^{u})

Substitute back

u = - 0.2 x

= 6 (- 5 e^{- 0.2 x})

Simplify

= - 30 e^{- 0.2 x}

Add a constant to the solution

= - 30 e^{- 0.2 x} + C

5 - x, 25 - x^{2}

\int \frac{1}{x ^{2}} arctan (x) d x

x \to 4 + lim (x - 4)

f (x) = 0.6 x^{1.5}

\frac{d y}{d x} = 9.8 - 0.196 y

What is the integral of (6e^{-0.2x}) ?
The integral of (6e^{-0.2x}) is -30e^{-0.2x}+C