What is the integral from 0 to 1 of (1-x^2)^{11} ?

The integral from 0 to 1 of (1-x^2)^{11} is 524288/2028117

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integral from 0 to 1 of (1-x^2)^{11}

\int_{0}^{1} (1 - x^{2})^{11} d x

\frac{524288}{2028117}

Decimal Notation

0.25850 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{11} d x

Expand

(1 - x^{2})^{11} : 1 - 11 x^{2} + 55 x^{4} - 165 x^{6} + 330 x^{8} - 462 x^{10} + 462 x^{12} - 330 x^{14} + 165 x^{16} - 55 x^{18} + 11 x^{20} - x^{22}

= \int_{0}^{1} 1 - 11 x^{2} + 55 x^{4} - 165 x^{6} + 330 x^{8} - 462 x^{10} + 462 x^{12} - 330 x^{14} + 165 x^{16} - 55 x^{18} + 11 x^{20} - x^{22} 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} 1 d x - \int_{0}^{1} 11 x^{2} d x + \int_{0}^{1} 55 x^{4} d x - \int_{0}^{1} 165 x^{6} d x + \int_{0}^{1} 330 x^{8} d x - \int_{0}^{1} 462 x^{10} d x + \int_{0}^{1} 462 x^{12} d x - \int_{0}^{1} 330 x^{14} d x + \int_{0}^{1} 165 x^{16} d x - \int_{0}^{1} 55 x^{18} d x + \int_{0}^{1} 11 x^{20} d x - \int_{0}^{1} x^{22} d x

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

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

\int_{0}^{1} 55 x^{4} d x = 11

\int_{0}^{1} 165 x^{6} d x = \frac{165}{7}

\int_{0}^{1} 330 x^{8} d x = \frac{110}{3}

\int_{0}^{1} 462 x^{10} d x = 42

\int_{0}^{1} 462 x^{12} d x = \frac{462}{13}

\int_{0}^{1} 330 x^{14} d x = 22

\int_{0}^{1} 165 x^{16} d x = \frac{165}{17}

\int_{0}^{1} 55 x^{18} d x = \frac{55}{19}

\int_{0}^{1} 11 x^{20} d x = \frac{11}{21}

\int_{0}^{1} x^{22} d x = \frac{1}{23}

= 1 - \frac{11}{3} + 11 - \frac{165}{7} + \frac{110}{3} - 42 + \frac{462}{13} - 22 + \frac{165}{17} - \frac{55}{19} + \frac{11}{21} - \frac{1}{23}

Simplify

1 - \frac{11}{3} + 11 - \frac{165}{7} + \frac{110}{3} - 42 + \frac{462}{13} - 22 + \frac{165}{17} - \frac{55}{19} + \frac{11}{21} - \frac{1}{23} : \frac{524288}{2028117}

= \frac{524288}{2028117}

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What is the integral from 0 to 1 of (1-x^2)^{11} ?
The integral from 0 to 1 of (1-x^2)^{11} is 524288/2028117