What is the integral from 0 to 1 of 6x^2(3-2x^3)^3 ?

The integral from 0 to 1 of 6x^2(3-2x^3)^3 is 20

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integral from 0 to 1 of 6x^2(3-2x^3)^3

\int_{0}^{1} 6 x^{2} (3 - 2 x^{3})^{3} d x

20

Solution steps

\int_{0}^{1} 6 x^{2} (3 - 2 x^{3})^{3} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} x^{2} (3 - 2 x^{3})^{3} d x

Expand

x^{2} (3 - 2 x^{3})^{3} : 27 x^{2} - 54 x^{5} + 36 x^{8} - 8 x^{11}

= 6 \cdot \int_{0}^{1} 27 x^{2} - 54 x^{5} + 36 x^{8} - 8 x^{11} 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

= 6 (\int_{0}^{1} 27 x^{2} d x - \int_{0}^{1} 54 x^{5} d x + \int_{0}^{1} 36 x^{8} d x - \int_{0}^{1} 8 x^{11} d x)

\int_{0}^{1} 27 x^{2} d x = 9

\int_{0}^{1} 54 x^{5} d x = 9

\int_{0}^{1} 36 x^{8} d x = 4

\int_{0}^{1} 8 x^{11} d x = \frac{2}{3}

= 6 (9 - 9 + 4 - \frac{2}{3})

Simplify

6 (9 - 9 + 4 - \frac{2}{3}) : 20

= 20

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What is the integral from 0 to 1 of 6x^2(3-2x^3)^3 ?
The integral from 0 to 1 of 6x^2(3-2x^3)^3 is 20