What is the integral from 0 to 1 of (1-x^4)^5 ?

The integral from 0 to 1 of (1-x^4)^5 is 8192/13923

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integral from 0 to 1 of (1-x^4)^5

\int_{0}^{1} (1 - x^{4})^{5} d x

\frac{8192}{13923}

Decimal Notation

0.58837 \dots

Solution steps

\int_{0}^{1} (1 - x^{4})^{5} d x

Expand

(1 - x^{4})^{5} : 1 - 5 x^{4} + 10 x^{8} - 10 x^{12} + 5 x^{16} - x^{20}

= \int_{0}^{1} 1 - 5 x^{4} + 10 x^{8} - 10 x^{12} + 5 x^{16} - x^{20} 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} 5 x^{4} d x + \int_{0}^{1} 10 x^{8} d x - \int_{0}^{1} 10 x^{12} d x + \int_{0}^{1} 5 x^{16} d x - \int_{0}^{1} x^{20} d x

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

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

\int_{0}^{1} 10 x^{8} d x = \frac{10}{9}

\int_{0}^{1} 10 x^{12} d x = \frac{10}{13}

\int_{0}^{1} 5 x^{16} d x = \frac{5}{17}

\int_{0}^{1} x^{20} d x = \frac{1}{21}

= 1 - 1 + \frac{10}{9} - \frac{10}{13} + \frac{5}{17} - \frac{1}{21}

Simplify

1 - 1 + \frac{10}{9} - \frac{10}{13} + \frac{5}{17} - \frac{1}{21} : \frac{8192}{13923}

= \frac{8192}{13923}

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

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What is the integral from 0 to 1 of (1-x^4)^5 ?
The integral from 0 to 1 of (1-x^4)^5 is 8192/13923