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

The integral from 0 to 1 of x(1-x)^5 is 1/42

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

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

\frac{1}{42}

Decimal

0.02380 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{2} x^{2} (1 - x)^{5} - \int - \frac{5}{2} x^{2} (1 - x)^{4} d x]_{0}^{1}

= [\frac{1}{2} x^{2} (1 - x)^{5} - \int - \frac{5}{2} x^{2} (1 - x)^{4} d x]_{0}^{1}

\int - \frac{5}{2} x^{2} (1 - x)^{4} d x = - \frac{1}{42} (35 x^{3} - 105 x^{4} + 126 x^{5} - 70 x^{6} + 15 x^{7})

= [\frac{1}{2} x^{2} (1 - x)^{5} - (- \frac{1}{42} (35 x^{3} - 105 x^{4} + 126 x^{5} - 70 x^{6} + 15 x^{7}))]_{0}^{1}

Simplify

= [\frac{1}{2} x^{2} (1 - x)^{5} + \frac{1}{42} (35 x^{3} - 105 x^{4} + 126 x^{5} - 70 x^{6} + 15 x^{7})]_{0}^{1}

Compute the boundaries:

\frac{1}{42}

= \frac{1}{42}

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What is the integral from 0 to 1 of x(1-x)^5 ?
The integral from 0 to 1 of x(1-x)^5 is 1/42