What is the integral from 0 to 12 of 1/36 (x(12-x)) ?

The integral from 0 to 12 of 1/36 (x(12-x)) is 8

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integral from 0 to 12 of 1/36 (x(12-x))

\int_{0}^{12} \frac{1}{36} (x (12 - x)) d x

8

Solution steps

\int_{0}^{12} \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_{0}^{12} x (12 - x) d x

Expand

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

= \frac{1}{36} \cdot \int_{0}^{12} 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_{0}^{12} 12 x d x - \int_{0}^{12} x^{2} d x)

\int_{0}^{12} 12 x d x = 864

\int_{0}^{12} x^{2} d x = 576

= \frac{1}{36} (864 - 576)

Simplify

\frac{1}{36} (864 - 576) : 8

= 8

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What is the integral from 0 to 12 of 1/36 (x(12-x)) ?
The integral from 0 to 12 of 1/36 (x(12-x)) is 8