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

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

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

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

8

Solution steps

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

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} t (12 - t) d t

Expand

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

= \frac{1}{36} \cdot \int_{0}^{12} 12 t - t^{2} d t

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 t d t - \int_{0}^{12} t^{2} d t)

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

\int_{0}^{12} t^{2} d t = 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 (t(12-t)) ?
The integral from 0 to 12 of 1/36 (t(12-t)) is 8