What is the integral from 0 to 1 of (1-x^4)^9 ?

The integral from 0 to 1 of (1-x^4)^9 is 33554432/65211575

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integral from 0 to 1 of (1-x^4)^9

\int_{0}^{1} (1 - x^{4})^{9} d x

\frac{33554432}{65211575}

Decimal Notation

0.51454 \dots

Solution steps

\int_{0}^{1} (1 - x^{4})^{9} d x

Expand

(1 - x^{4})^{9} : 1 - 9 x^{4} + 36 x^{8} - 84 x^{12} + 126 x^{16} - 126 x^{20} + 84 x^{24} - 36 x^{28} + 9 x^{32} - x^{36}

= \int_{0}^{1} 1 - 9 x^{4} + 36 x^{8} - 84 x^{12} + 126 x^{16} - 126 x^{20} + 84 x^{24} - 36 x^{28} + 9 x^{32} - x^{36} 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^{4} d x + \int_{0}^{1} 36 x^{8} d x - \int_{0}^{1} 84 x^{12} d x + \int_{0}^{1} 126 x^{16} d x - \int_{0}^{1} 126 x^{20} d x + \int_{0}^{1} 84 x^{24} d x - \int_{0}^{1} 36 x^{28} d x + \int_{0}^{1} 9 x^{32} d x - \int_{0}^{1} x^{36} d x

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

\int_{0}^{1} 9 x^{4} d x = \frac{9}{5}

\int_{0}^{1} 36 x^{8} d x = 4

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

\int_{0}^{1} 126 x^{16} d x = \frac{126}{17}

\int_{0}^{1} 126 x^{20} d x = 6

\int_{0}^{1} 84 x^{24} d x = \frac{84}{25}

\int_{0}^{1} 36 x^{28} d x = \frac{36}{29}

\int_{0}^{1} 9 x^{32} d x = \frac{3}{11}

\int_{0}^{1} x^{36} d x = \frac{1}{37}

= 1 - \frac{9}{5} + 4 - \frac{84}{13} + \frac{126}{17} - 6 + \frac{84}{25} - \frac{36}{29} + \frac{3}{11} - \frac{1}{37}

Simplify

1 - \frac{9}{5} + 4 - \frac{84}{13} + \frac{126}{17} - 6 + \frac{84}{25} - \frac{36}{29} + \frac{3}{11} - \frac{1}{37} : \frac{33554432}{65211575}

= \frac{33554432}{65211575}

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What is the integral from 0 to 1 of (1-x^4)^9 ?
The integral from 0 to 1 of (1-x^4)^9 is 33554432/65211575