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

The integral from 0 to 1 of x^4(1-x)^4 is 1/630

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

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

\frac{1}{630}

Decimal

0.00158 \dots

Solution steps

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

Expand

x^{4} (1 - x)^{4} : x^{4} - 4 x^{5} + 6 x^{6} - 4 x^{7} + x^{8}

= \int_{0}^{1} x^{4} - 4 x^{5} + 6 x^{6} - 4 x^{7} + x^{8} 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} x^{4} d x - \int_{0}^{1} 4 x^{5} d x + \int_{0}^{1} 6 x^{6} d x - \int_{0}^{1} 4 x^{7} d x + \int_{0}^{1} x^{8} d x

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

\int_{0}^{1} 4 x^{5} d x = \frac{2}{3}

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

\int_{0}^{1} 4 x^{7} d x = \frac{1}{2}

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

= \frac{1}{5} - \frac{2}{3} + \frac{6}{7} - \frac{1}{2} + \frac{1}{9}

Simplify

\frac{1}{5} - \frac{2}{3} + \frac{6}{7} - \frac{1}{2} + \frac{1}{9} : \frac{1}{630}

= \frac{1}{630}

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What is the integral from 0 to 1 of x^4(1-x)^4 ?
The integral from 0 to 1 of x^4(1-x)^4 is 1/630