What is the integral of (3z-6)/3 ?

The integral of (3z-6)/3 is 1/2 z(z-4)+C

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integral of (3z-6)/3

\int \frac{3 z - 6}{3} d z

\frac{1}{2} z (z - 4) + C

Solution steps

\int \frac{3 z - 6}{3} d z

Take the constant out:

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

= \frac{1}{3} \cdot \int 3 z - 6 d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 3 z d z - \int 6 d z)

\int 3 z d z = \frac{3 z ^{2}}{2}

\int 6 d z = 6 z

= \frac{1}{3} (\frac{3 z ^{2}}{2} - 6 z)

Simplify

\frac{1}{3} (\frac{3 z ^{2}}{2} - 6 z) : \frac{1}{2} z (z - 4)

= \frac{1}{2} z (z - 4)

Add a constant to the solution

= \frac{1}{2} z (z - 4) + C

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