What is the integral from 0 to 6 of (36x-x^3)/(144) ?

The integral from 0 to 6 of (36x-x^3)/(144) is 9/4

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integral from 0 to 6 of (36x-x^3)/(144)

\int_{0}^{6} \frac{36 x - x ^{3}}{144} d x

\frac{9}{4}

Decimal

2.25

Solution steps

\int_{0}^{6} \frac{36 x - x ^{3}}{144} d x

Take the constant out:

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

= \frac{1}{144} \cdot \int_{0}^{6} 36 x - x^{3} 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}{144} (\int_{0}^{6} 36 x d x - \int_{0}^{6} x^{3} d x)

\int_{0}^{6} 36 x d x = 648

\int_{0}^{6} x^{3} d x = 324

= \frac{1}{144} (648 - 324)

Simplify

\frac{1}{144} (648 - 324) : \frac{9}{4}

= \frac{9}{4}

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What is the integral from 0 to 6 of (36x-x^3)/(144) ?
The integral from 0 to 6 of (36x-x^3)/(144) is 9/4