What is the integral from 0 to 1 of x^3(1-x)^2 ?

The integral from 0 to 1 of x^3(1-x)^2 is 1/60

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integral from 0 to 1 of x^3(1-x)^2

\int_{0}^{1} x^{3} (1 - x)^{2} d x

\frac{1}{60}

Decimal

0.01666 \dots

Solution steps

\int_{0}^{1} x^{3} (1 - x)^{2} d x

Apply Integration By Parts

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

\int - \frac{1}{2} x^{4} (- x + 1) d x = - \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5})

= [\frac{1}{4} x^{4} (1 - x)^{2} - (- \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5}))]_{0}^{1}

Simplify

[\frac{1}{4} x^{4} (1 - x)^{2} - (- \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5}))]_{0}^{1} : [\frac{1}{60} (10 x^{6} - 24 x^{5} + 15 x^{4})]_{0}^{1}

= [\frac{1}{60} (10 x^{6} - 24 x^{5} + 15 x^{4})]_{0}^{1}

Compute the boundaries:

\frac{1}{60}

= \frac{1}{60}

What is the integral from 0 to 1 of x^3(1-x)^2 ?
The integral from 0 to 1 of x^3(1-x)^2 is 1/60