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

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

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

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

\frac{1}{60}

Decimal

0.01666 \dots

Solution steps

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

Expand

x^{2} (1 - x)^{3} : x^{2} - 3 x^{3} + 3 x^{4} - x^{5}

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

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

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

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

\int_{0}^{1} x^{5} d x = \frac{1}{6}

= \frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6}

Simplify

\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6} : \frac{1}{60}

= \frac{1}{60}

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