What is the integral from 0 to 1 of (1-x^4)^7 ?

The integral from 0 to 1 of (1-x^4)^7 is 262144/480675

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integral from 0 to 1 of (1-x^4)^7

\int_{0}^{1} (1 - x^{4})^{7} d x

\frac{262144}{480675}

Decimal

0.54536 \dots

Solution steps

\int_{0}^{1} (1 - x^{4})^{7} d x

Expand

(1 - x^{4})^{7} : 1 - 7 x^{4} + 21 x^{8} - 35 x^{12} + 35 x^{16} - 21 x^{20} + 7 x^{24} - x^{28}

= \int_{0}^{1} 1 - 7 x^{4} + 21 x^{8} - 35 x^{12} + 35 x^{16} - 21 x^{20} + 7 x^{24} - x^{28} 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} 7 x^{4} d x + \int_{0}^{1} 21 x^{8} d x - \int_{0}^{1} 35 x^{12} d x + \int_{0}^{1} 35 x^{16} d x - \int_{0}^{1} 21 x^{20} d x + \int_{0}^{1} 7 x^{24} d x - \int_{0}^{1} x^{28} d x

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

\int_{0}^{1} 7 x^{4} d x = \frac{7}{5}

\int_{0}^{1} 21 x^{8} d x = \frac{7}{3}

\int_{0}^{1} 35 x^{12} d x = \frac{35}{13}

\int_{0}^{1} 35 x^{16} d x = \frac{35}{17}

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

\int_{0}^{1} 7 x^{24} d x = \frac{7}{25}

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

= 1 - \frac{7}{5} + \frac{7}{3} - \frac{35}{13} + \frac{35}{17} - 1 + \frac{7}{25} - \frac{1}{29}

Simplify

1 - \frac{7}{5} + \frac{7}{3} - \frac{35}{13} + \frac{35}{17} - 1 + \frac{7}{25} - \frac{1}{29} : \frac{262144}{480675}

= \frac{262144}{480675}

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What is the integral from 0 to 1 of (1-x^4)^7 ?
The integral from 0 to 1 of (1-x^4)^7 is 262144/480675