What is the integral of 1/36 (x(12-x)) ?

The integral of 1/36 (x(12-x)) is 1/36 (6x^2-(x^3)/3)+C

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integral of 1/36 (x(12-x))

\int \frac{1}{36} (x (12 - x)) d x

\frac{1}{36} (6 x^{2} - \frac{x ^{3}}{3}) + C

Solution steps

\int \frac{1}{36} x (12 - x) d x

Take the constant out:

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

= \frac{1}{36} \cdot \int x (12 - x) d x

Expand

x (12 - x) : 12 x - x^{2}

= \frac{1}{36} \cdot \int 12 x - x^{2} d x

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}{36} (\int 12 x d x - \int x^{2} d x)

\int 12 x d x = 6 x^{2}

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

= \frac{1}{36} (6 x^{2} - \frac{x ^{3}}{3})

Add a constant to the solution

= \frac{1}{36} (6 x^{2} - \frac{x ^{3}}{3}) + C

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What is the integral of 1/36 (x(12-x)) ?
The integral of 1/36 (x(12-x)) is 1/36 (6x^2-(x^3)/3)+C