What is the integral from 0 to 1 of (1-x^4)^6 ?

The integral from 0 to 1 of (1-x^4)^6 is 65536/116025

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integral from 0 to 1 of (1-x^4)^6

\int_{0}^{1} (1 - x^{4})^{6} d x

\frac{65536}{116025}

Decimal Notation

0.56484 \dots

Solution steps

\int_{0}^{1} (1 - x^{4})^{6} d x

Expand

(1 - x^{4})^{6} : 1 - 6 x^{4} + 15 x^{8} - 20 x^{12} + 15 x^{16} - 6 x^{20} + x^{24}

= \int_{0}^{1} 1 - 6 x^{4} + 15 x^{8} - 20 x^{12} + 15 x^{16} - 6 x^{20} + x^{24} 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} 6 x^{4} d x + \int_{0}^{1} 15 x^{8} d x - \int_{0}^{1} 20 x^{12} d x + \int_{0}^{1} 15 x^{16} d x - \int_{0}^{1} 6 x^{20} d x + \int_{0}^{1} x^{24} d x

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

\int_{0}^{1} 6 x^{4} d x = \frac{6}{5}

\int_{0}^{1} 15 x^{8} d x = \frac{5}{3}

\int_{0}^{1} 20 x^{12} d x = \frac{20}{13}

\int_{0}^{1} 15 x^{16} d x = \frac{15}{17}

\int_{0}^{1} 6 x^{20} d x = \frac{2}{7}

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

= 1 - \frac{6}{5} + \frac{5}{3} - \frac{20}{13} + \frac{15}{17} - \frac{2}{7} + \frac{1}{25}

Simplify

1 - \frac{6}{5} + \frac{5}{3} - \frac{20}{13} + \frac{15}{17} - \frac{2}{7} + \frac{1}{25} : \frac{65536}{116025}

= \frac{65536}{116025}

\int_{0}^{2} (x^{4} + 5)^{3} x^{3} d x

\int_{0}^{\infty} \frac{x}{x ^{2} + 4} d x

\int_{0}^{1} (x + 5)^{3} d x

\int_{0}^{1} (1 - x^{2})^{6} d x

\int_{1}^{\infty} \frac{x}{x ^{2} + 9} d x

What is the integral from 0 to 1 of (1-x^4)^6 ?
The integral from 0 to 1 of (1-x^4)^6 is 65536/116025