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

sin (3 x), 0

12 (t + 1) (\frac{d y}{d t}) - 8 y = 32 t, y (0) = 2

y = \frac{1}{t ^{3} + 2 t ^{2} - 1}

x = - 5, x = 1, y = 3 x, y = x^{2} - 4

\frac{d}{d x} (\frac{4 x - 7}{2 x + 1})

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