What is the integral from 0 to 1 of x^230x^2(1-x)^2 ?

The integral from 0 to 1 of x^230x^2(1-x)^2 is 2/7

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integral from 0 to 1 of x^230x^2(1-x)^2

\int_{0}^{1} x^{2} 30 x^{2} (1 - x)^{2} d x

\frac{2}{7}

Decimal

0.28571 \dots

Solution steps

\int_{0}^{1} x^{2} \cdot 30 x^{2} (1 - x)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 30 \cdot \int_{0}^{1} x^{2} x^{2} (1 - x)^{2} d x

Simplify

x^{2} x^{2} : x^{4}

= 30 \cdot \int_{0}^{1} x^{4} (1 - x)^{2} d x

Expand

x^{4} (1 - x)^{2} : x^{4} - 2 x^{5} + x^{6}

= 30 \cdot \int_{0}^{1} x^{4} - 2 x^{5} + x^{6} 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

= 30 (\int_{0}^{1} x^{4} d x - \int_{0}^{1} 2 x^{5} d x + \int_{0}^{1} x^{6} d x)

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

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

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

= 30 (\frac{1}{5} - \frac{1}{3} + \frac{1}{7})

Simplify

30 (\frac{1}{5} - \frac{1}{3} + \frac{1}{7}) : \frac{2}{7}

= \frac{2}{7}

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What is the integral from 0 to 1 of x^230x^2(1-x)^2 ?
The integral from 0 to 1 of x^230x^2(1-x)^2 is 2/7