What is the integral from 0 to 1 of (1-x^2)^7 ?

The integral from 0 to 1 of (1-x^2)^7 is 2048/6435

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integral from 0 to 1 of (1-x^2)^7

\int_{0}^{1} (1 - x^{2})^{7} d x

\frac{2048}{6435}

Decimal Notation

0.31825 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{7} d x

Expand

(1 - x^{2})^{7} : 1 - 7 x^{2} + 21 x^{4} - 35 x^{6} + 35 x^{8} - 21 x^{10} + 7 x^{12} - x^{14}

= \int_{0}^{1} 1 - 7 x^{2} + 21 x^{4} - 35 x^{6} + 35 x^{8} - 21 x^{10} + 7 x^{12} - x^{14} 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} 7 x^{2} d x + \int_{0}^{1} 21 x^{4} d x - \int_{0}^{1} 35 x^{6} d x + \int_{0}^{1} 35 x^{8} d x - \int_{0}^{1} 21 x^{10} d x + \int_{0}^{1} 7 x^{12} d x - \int_{0}^{1} x^{14} d x

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

\int_{0}^{1} 7 x^{2} d x = \frac{7}{3}

\int_{0}^{1} 21 x^{4} d x = \frac{21}{5}

\int_{0}^{1} 35 x^{6} d x = 5

\int_{0}^{1} 35 x^{8} d x = \frac{35}{9}

\int_{0}^{1} 21 x^{10} d x = \frac{21}{11}

\int_{0}^{1} 7 x^{12} d x = \frac{7}{13}

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

= 1 - \frac{7}{3} + \frac{21}{5} - 5 + \frac{35}{9} - \frac{21}{11} + \frac{7}{13} - \frac{1}{15}

Simplify

1 - \frac{7}{3} + \frac{21}{5} - 5 + \frac{35}{9} - \frac{21}{11} + \frac{7}{13} - \frac{1}{15} : \frac{2048}{6435}

= \frac{2048}{6435}

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What is the integral from 0 to 1 of (1-x^2)^7 ?
The integral from 0 to 1 of (1-x^2)^7 is 2048/6435