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

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

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

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

\frac{2}{7}

Decimal

0.28571 \dots

Solution steps

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

Take the constant out:

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

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

Simplify

y^{2} y^{2} : y^{4}

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

Expand

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

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

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} y^{4} d y - \int_{0}^{1} 2 y^{5} d y + \int_{0}^{1} y^{6} d y)

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

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

\int_{0}^{1} y^{6} d y = \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 y^230y^2(1-y)^2 ?
The integral from 0 to 1 of y^230y^2(1-y)^2 is 2/7