What is the integral from 0 to 1 of x(1-x^2)^9 ?

The integral from 0 to 1 of x(1-x^2)^9 is 1/20

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integral from 0 to 1 of x(1-x^2)^9

\int_{0}^{1} x (1 - x^{2})^{9} d x

\frac{1}{20}

Decimal

0.05

Solution steps

\int_{0}^{1} x (1 - x^{2})^{9} d x

Expand

x (1 - x^{2})^{9} : x - 9 x^{3} + 36 x^{5} - 84 x^{7} + 126 x^{9} - 126 x^{11} + 84 x^{13} - 36 x^{15} + 9 x^{17} - x^{19}

= \int_{0}^{1} x - 9 x^{3} + 36 x^{5} - 84 x^{7} + 126 x^{9} - 126 x^{11} + 84 x^{13} - 36 x^{15} + 9 x^{17} - x^{19} 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} x d x - \int_{0}^{1} 9 x^{3} d x + \int_{0}^{1} 36 x^{5} d x - \int_{0}^{1} 84 x^{7} d x + \int_{0}^{1} 126 x^{9} d x - \int_{0}^{1} 126 x^{11} d x + \int_{0}^{1} 84 x^{13} d x - \int_{0}^{1} 36 x^{15} d x + \int_{0}^{1} 9 x^{17} d x - \int_{0}^{1} x^{19} d x

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

\int_{0}^{1} 9 x^{3} d x = \frac{9}{4}

\int_{0}^{1} 36 x^{5} d x = 6

\int_{0}^{1} 84 x^{7} d x = \frac{21}{2}

\int_{0}^{1} 126 x^{9} d x = \frac{63}{5}

\int_{0}^{1} 126 x^{11} d x = \frac{21}{2}

\int_{0}^{1} 84 x^{13} d x = 6

\int_{0}^{1} 36 x^{15} d x = \frac{9}{4}

\int_{0}^{1} 9 x^{17} d x = \frac{1}{2}

\int_{0}^{1} x^{19} d x = \frac{1}{20}

= \frac{1}{2} - \frac{9}{4} + 6 - \frac{21}{2} + \frac{63}{5} - \frac{21}{2} + 6 - \frac{9}{4} + \frac{1}{2} - \frac{1}{20}

Simplify

\frac{1}{2} - \frac{9}{4} + 6 - \frac{21}{2} + \frac{63}{5} - \frac{21}{2} + 6 - \frac{9}{4} + \frac{1}{2} - \frac{1}{20} : \frac{1}{20}

= \frac{1}{20}

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What is the integral from 0 to 1 of x(1-x^2)^9 ?
The integral from 0 to 1 of x(1-x^2)^9 is 1/20