What is the integral from 0 to 1 of (1-x^2)^{10} ?

The integral from 0 to 1 of (1-x^2)^{10} is 262144/969969

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integral from 0 to 1 of (1-x^2)^{10}

\int_{0}^{1} (1 - x^{2})^{10} d x

\frac{262144}{969969}

Decimal Notation

0.27026 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{10} d x

Expand

(1 - x^{2})^{10} : 1 - 10 x^{2} + 45 x^{4} - 120 x^{6} + 210 x^{8} - 252 x^{10} + 210 x^{12} - 120 x^{14} + 45 x^{16} - 10 x^{18} + x^{20}

= \int_{0}^{1} 1 - 10 x^{2} + 45 x^{4} - 120 x^{6} + 210 x^{8} - 252 x^{10} + 210 x^{12} - 120 x^{14} + 45 x^{16} - 10 x^{18} + x^{20} 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} 10 x^{2} d x + \int_{0}^{1} 45 x^{4} d x - \int_{0}^{1} 120 x^{6} d x + \int_{0}^{1} 210 x^{8} d x - \int_{0}^{1} 252 x^{10} d x + \int_{0}^{1} 210 x^{12} d x - \int_{0}^{1} 120 x^{14} d x + \int_{0}^{1} 45 x^{16} d x - \int_{0}^{1} 10 x^{18} d x + \int_{0}^{1} x^{20} d x

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

\int_{0}^{1} 10 x^{2} d x = \frac{10}{3}

\int_{0}^{1} 45 x^{4} d x = 9

\int_{0}^{1} 120 x^{6} d x = \frac{120}{7}

\int_{0}^{1} 210 x^{8} d x = \frac{70}{3}

\int_{0}^{1} 252 x^{10} d x = \frac{252}{11}

\int_{0}^{1} 210 x^{12} d x = \frac{210}{13}

\int_{0}^{1} 120 x^{14} d x = 8

\int_{0}^{1} 45 x^{16} d x = \frac{45}{17}

\int_{0}^{1} 10 x^{18} d x = \frac{10}{19}

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

= 1 - \frac{10}{3} + 9 - \frac{120}{7} + \frac{70}{3} - \frac{252}{11} + \frac{210}{13} - 8 + \frac{45}{17} - \frac{10}{19} + \frac{1}{21}

Simplify

1 - \frac{10}{3} + 9 - \frac{120}{7} + \frac{70}{3} - \frac{252}{11} + \frac{210}{13} - 8 + \frac{45}{17} - \frac{10}{19} + \frac{1}{21} : \frac{262144}{969969}

= \frac{262144}{969969}

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What is the integral from 0 to 1 of (1-x^2)^{10} ?
The integral from 0 to 1 of (1-x^2)^{10} is 262144/969969