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

The integral from 0 to 1 of (1-x^2)^9 is 65536/230945

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

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

\frac{65536}{230945}

Decimal

0.28377 \dots

Solution steps

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

Expand

(1 - x^{2})^{9} : 1 - 9 x^{2} + 36 x^{4} - 84 x^{6} + 126 x^{8} - 126 x^{10} + 84 x^{12} - 36 x^{14} + 9 x^{16} - x^{18}

= \int_{0}^{1} 1 - 9 x^{2} + 36 x^{4} - 84 x^{6} + 126 x^{8} - 126 x^{10} + 84 x^{12} - 36 x^{14} + 9 x^{16} - x^{18} 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} 9 x^{2} d x + \int_{0}^{1} 36 x^{4} d x - \int_{0}^{1} 84 x^{6} d x + \int_{0}^{1} 126 x^{8} d x - \int_{0}^{1} 126 x^{10} d x + \int_{0}^{1} 84 x^{12} d x - \int_{0}^{1} 36 x^{14} d x + \int_{0}^{1} 9 x^{16} d x - \int_{0}^{1} x^{18} d x

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

\int_{0}^{1} 9 x^{2} d x = 3

\int_{0}^{1} 36 x^{4} d x = \frac{36}{5}

\int_{0}^{1} 84 x^{6} d x = 12

\int_{0}^{1} 126 x^{8} d x = 14

\int_{0}^{1} 126 x^{10} d x = \frac{126}{11}

\int_{0}^{1} 84 x^{12} d x = \frac{84}{13}

\int_{0}^{1} 36 x^{14} d x = \frac{12}{5}

\int_{0}^{1} 9 x^{16} d x = \frac{9}{17}

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

= 1 - 3 + \frac{36}{5} - 12 + 14 - \frac{126}{11} + \frac{84}{13} - \frac{12}{5} + \frac{9}{17} - \frac{1}{19}

Simplify

1 - 3 + \frac{36}{5} - 12 + 14 - \frac{126}{11} + \frac{84}{13} - \frac{12}{5} + \frac{9}{17} - \frac{1}{19} : \frac{65536}{230945}

= \frac{65536}{230945}

What is the integral from 0 to 1 of (1-x^2)^9 ?
The integral from 0 to 1 of (1-x^2)^9 is 65536/230945