What is the integral from 0 to 1 of (1-x^4)^{10} ?

The integral from 0 to 1 of (1-x^4)^{10} is 268435456/534734915

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integral from 0 to 1 of (1-x^4)^{10}

\int_{0}^{1} (1 - x^{4})^{10} d x

\frac{268435456}{534734915}

Decimal Notation

0.50199 \dots

Solution steps

\int_{0}^{1} (1 - x^{4})^{10} d x

Expand

(1 - x^{4})^{10} : 1 - 10 x^{4} + 45 x^{8} - 120 x^{12} + 210 x^{16} - 252 x^{20} + 210 x^{24} - 120 x^{28} + 45 x^{32} - 10 x^{36} + x^{40}

= \int_{0}^{1} 1 - 10 x^{4} + 45 x^{8} - 120 x^{12} + 210 x^{16} - 252 x^{20} + 210 x^{24} - 120 x^{28} + 45 x^{32} - 10 x^{36} + x^{40} 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} 10 x^{4} d x + \int_{0}^{1} 45 x^{8} d x - \int_{0}^{1} 120 x^{12} d x + \int_{0}^{1} 210 x^{16} d x - \int_{0}^{1} 252 x^{20} d x + \int_{0}^{1} 210 x^{24} d x - \int_{0}^{1} 120 x^{28} d x + \int_{0}^{1} 45 x^{32} d x - \int_{0}^{1} 10 x^{36} d x + \int_{0}^{1} x^{40} d x

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

\int_{0}^{1} 10 x^{4} d x = 2

\int_{0}^{1} 45 x^{8} d x = 5

\int_{0}^{1} 120 x^{12} d x = \frac{120}{13}

\int_{0}^{1} 210 x^{16} d x = \frac{210}{17}

\int_{0}^{1} 252 x^{20} d x = 12

\int_{0}^{1} 210 x^{24} d x = \frac{42}{5}

\int_{0}^{1} 120 x^{28} d x = \frac{120}{29}

\int_{0}^{1} 45 x^{32} d x = \frac{15}{11}

\int_{0}^{1} 10 x^{36} d x = \frac{10}{37}

\int_{0}^{1} x^{40} d x = \frac{1}{41}

= 1 - 2 + 5 - \frac{120}{13} + \frac{210}{17} - 12 + \frac{42}{5} - \frac{120}{29} + \frac{15}{11} - \frac{10}{37} + \frac{1}{41}

Simplify

1 - 2 + 5 - \frac{120}{13} + \frac{210}{17} - 12 + \frac{42}{5} - \frac{120}{29} + \frac{15}{11} - \frac{10}{37} + \frac{1}{41} : \frac{268435456}{534734915}

= \frac{268435456}{534734915}

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\int_{1}^{e} (ln (x))^{10} d x

What is the integral from 0 to 1 of (1-x^4)^{10} ?
The integral from 0 to 1 of (1-x^4)^{10} is 268435456/534734915