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

The integral from 0 to 1 of (1-x^4)^4 is 2048/3315

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

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

\frac{2048}{3315}

Decimal Notation

0.61779 \dots

Solution steps

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

Expand

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

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

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

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

\int_{0}^{1} 6 x^{8} d x = \frac{2}{3}

\int_{0}^{1} 4 x^{12} d x = \frac{4}{13}

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

= 1 - \frac{4}{5} + \frac{2}{3} - \frac{4}{13} + \frac{1}{17}

Simplify

1 - \frac{4}{5} + \frac{2}{3} - \frac{4}{13} + \frac{1}{17} : \frac{2048}{3315}

= \frac{2048}{3315}

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What is the integral from 0 to 1 of (1-x^4)^4 ?
The integral from 0 to 1 of (1-x^4)^4 is 2048/3315