What is the integral from 0 to 1 of (1-x^2)^8 ?

The integral from 0 to 1 of (1-x^2)^8 is 32768/109395

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integral from 0 to 1 of (1-x^2)^8

\int_{0}^{1} (1 - x^{2})^{8} d x

\frac{32768}{109395}

Decimal

0.29953 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{8} d x

Expand

(1 - x^{2})^{8} : 1 - 8 x^{2} + 28 x^{4} - 56 x^{6} + 70 x^{8} - 56 x^{10} + 28 x^{12} - 8 x^{14} + x^{16}

= \int_{0}^{1} 1 - 8 x^{2} + 28 x^{4} - 56 x^{6} + 70 x^{8} - 56 x^{10} + 28 x^{12} - 8 x^{14} + x^{16} 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} 8 x^{2} d x + \int_{0}^{1} 28 x^{4} d x - \int_{0}^{1} 56 x^{6} d x + \int_{0}^{1} 70 x^{8} d x - \int_{0}^{1} 56 x^{10} d x + \int_{0}^{1} 28 x^{12} d x - \int_{0}^{1} 8 x^{14} d x + \int_{0}^{1} x^{16} d x

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

\int_{0}^{1} 8 x^{2} d x = \frac{8}{3}

\int_{0}^{1} 28 x^{4} d x = \frac{28}{5}

\int_{0}^{1} 56 x^{6} d x = 8

\int_{0}^{1} 70 x^{8} d x = \frac{70}{9}

\int_{0}^{1} 56 x^{10} d x = \frac{56}{11}

\int_{0}^{1} 28 x^{12} d x = \frac{28}{13}

\int_{0}^{1} 8 x^{14} d x = \frac{8}{15}

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

= 1 - \frac{8}{3} + \frac{28}{5} - 8 + \frac{70}{9} - \frac{56}{11} + \frac{28}{13} - \frac{8}{15} + \frac{1}{17}

Simplify

1 - \frac{8}{3} + \frac{28}{5} - 8 + \frac{70}{9} - \frac{56}{11} + \frac{28}{13} - \frac{8}{15} + \frac{1}{17} : \frac{32768}{109395}

= \frac{32768}{109395}

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What is the integral from 0 to 1 of (1-x^2)^8 ?
The integral from 0 to 1 of (1-x^2)^8 is 32768/109395