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

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

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

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

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

Solution steps

\int \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 t (12 - t) d t

Expand

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

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

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

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

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

Add a constant to the solution

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

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