What is the integral from 0 to 1 of x(1-x)^9 ?

The integral from 0 to 1 of x(1-x)^9 is 1/110

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integral from 0 to 1 of x(1-x)^9

\int_{0}^{1} x (1 - x)^{9} d x

\frac{1}{110}

Decimal

0.00909 \dots

Solution steps

\int_{0}^{1} x (1 - x)^{9} d x

Expand

x (1 - x)^{9} : x - 9 x^{2} + 36 x^{3} - 84 x^{4} + 126 x^{5} - 126 x^{6} + 84 x^{7} - 36 x^{8} + 9 x^{9} - x^{10}

= \int_{0}^{1} x - 9 x^{2} + 36 x^{3} - 84 x^{4} + 126 x^{5} - 126 x^{6} + 84 x^{7} - 36 x^{8} + 9 x^{9} - x^{10} 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 d x - \int_{0}^{1} 9 x^{2} d x + \int_{0}^{1} 36 x^{3} d x - \int_{0}^{1} 84 x^{4} d x + \int_{0}^{1} 126 x^{5} d x - \int_{0}^{1} 126 x^{6} d x + \int_{0}^{1} 84 x^{7} d x - \int_{0}^{1} 36 x^{8} d x + \int_{0}^{1} 9 x^{9} d x - \int_{0}^{1} x^{10} d x

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

\int_{0}^{1} 9 x^{2} d x = 3

\int_{0}^{1} 36 x^{3} d x = 9

\int_{0}^{1} 84 x^{4} d x = \frac{84}{5}

\int_{0}^{1} 126 x^{5} d x = 21

\int_{0}^{1} 126 x^{6} d x = 18

\int_{0}^{1} 84 x^{7} d x = \frac{21}{2}

\int_{0}^{1} 36 x^{8} d x = 4

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

\int_{0}^{1} x^{10} d x = \frac{1}{11}

= \frac{1}{2} - 3 + 9 - \frac{84}{5} + 21 - 18 + \frac{21}{2} - 4 + \frac{9}{10} - \frac{1}{11}

Simplify

\frac{1}{2} - 3 + 9 - \frac{84}{5} + 21 - 18 + \frac{21}{2} - 4 + \frac{9}{10} - \frac{1}{11} : \frac{1}{110}

= \frac{1}{110}

What is the integral from 0 to 1 of x(1-x)^9 ?
The integral from 0 to 1 of x(1-x)^9 is 1/110