What is the integral from 0 to 1 of (1-x^4)^8 ?

The integral from 0 to 1 of (1-x^4)^8 is 8388608/15862275

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integral from 0 to 1 of (1-x^4)^8

\int_{0}^{1} (1 - x^{4})^{8} d x

\frac{8388608}{15862275}

Decimal Notation

0.52884 \dots

Solution steps

\int_{0}^{1} (1 - x^{4})^{8} d x

Expand

(1 - x^{4})^{8} : 1 - 8 x^{4} + 28 x^{8} - 56 x^{12} + 70 x^{16} - 56 x^{20} + 28 x^{24} - 8 x^{28} + x^{32}

= \int_{0}^{1} 1 - 8 x^{4} + 28 x^{8} - 56 x^{12} + 70 x^{16} - 56 x^{20} + 28 x^{24} - 8 x^{28} + x^{32} 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^{4} d x + \int_{0}^{1} 28 x^{8} d x - \int_{0}^{1} 56 x^{12} d x + \int_{0}^{1} 70 x^{16} d x - \int_{0}^{1} 56 x^{20} d x + \int_{0}^{1} 28 x^{24} d x - \int_{0}^{1} 8 x^{28} d x + \int_{0}^{1} x^{32} d x

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

\int_{0}^{1} 8 x^{4} d x = \frac{8}{5}

\int_{0}^{1} 28 x^{8} d x = \frac{28}{9}

\int_{0}^{1} 56 x^{12} d x = \frac{56}{13}

\int_{0}^{1} 70 x^{16} d x = \frac{70}{17}

\int_{0}^{1} 56 x^{20} d x = \frac{8}{3}

\int_{0}^{1} 28 x^{24} d x = \frac{28}{25}

\int_{0}^{1} 8 x^{28} d x = \frac{8}{29}

\int_{0}^{1} x^{32} d x = \frac{1}{33}

= 1 - \frac{8}{5} + \frac{28}{9} - \frac{56}{13} + \frac{70}{17} - \frac{8}{3} + \frac{28}{25} - \frac{8}{29} + \frac{1}{33}

Simplify

1 - \frac{8}{5} + \frac{28}{9} - \frac{56}{13} + \frac{70}{17} - \frac{8}{3} + \frac{28}{25} - \frac{8}{29} + \frac{1}{33} : \frac{8388608}{15862275}

= \frac{8388608}{15862275}

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What is the integral from 0 to 1 of (1-x^4)^8 ?
The integral from 0 to 1 of (1-x^4)^8 is 8388608/15862275