What is the integral from 0 to 1 of x^2(4-x^3)^3 ?

The integral from 0 to 1 of x^2(4-x^3)^3 is 175/12

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integral from 0 to 1 of x^2(4-x^3)^3

\int_{0}^{1} x^{2} (4 - x^{3})^{3} d x

\frac{175}{12}

Decimal

14.58333 \dots

Solution steps

\int_{0}^{1} x^{2} (4 - x^{3})^{3} d x

Expand

x^{2} (4 - x^{3})^{3} : 64 x^{2} - 48 x^{5} + 12 x^{8} - x^{11}

= \int_{0}^{1} 64 x^{2} - 48 x^{5} + 12 x^{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

= \int_{0}^{1} 64 x^{2} d x - \int_{0}^{1} 48 x^{5} d x + \int_{0}^{1} 12 x^{8} d x - \int_{0}^{1} x^{11} d x

\int_{0}^{1} 64 x^{2} d x = \frac{64}{3}

\int_{0}^{1} 48 x^{5} d x = 8

\int_{0}^{1} 12 x^{8} d x = \frac{4}{3}

\int_{0}^{1} x^{11} d x = \frac{1}{12}

= \frac{64}{3} - 8 + \frac{4}{3} - \frac{1}{12}

Simplify

\frac{64}{3} - 8 + \frac{4}{3} - \frac{1}{12} : \frac{175}{12}

= \frac{175}{12}

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What is the integral from 0 to 1 of x^2(4-x^3)^3 ?
The integral from 0 to 1 of x^2(4-x^3)^3 is 175/12