What is the integral from 0 to 1 of cy^2(1-y)^4 ?

The integral from 0 to 1 of cy^2(1-y)^4 is 1/105 c

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integral from 0 to 1 of cy^2(1-y)^4

\int_{0}^{1} c y^{2} (1 - y)^{4} d y

\frac{1}{105} c

Solution steps

\int_{0}^{1} c y^{2} (1 - y)^{4} d y

Take the constant out:

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

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

Expand

y^{2} (1 - y)^{4} : y^{2} - 4 y^{3} + 6 y^{4} - 4 y^{5} + y^{6}

= c \cdot \int_{0}^{1} y^{2} - 4 y^{3} + 6 y^{4} - 4 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

= c (\int_{0}^{1} y^{2} d y - \int_{0}^{1} 4 y^{3} d y + \int_{0}^{1} 6 y^{4} d y - \int_{0}^{1} 4 y^{5} d y + \int_{0}^{1} y^{6} d y)

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

\int_{0}^{1} 4 y^{3} d y = 1

\int_{0}^{1} 6 y^{4} d y = \frac{6}{5}

\int_{0}^{1} 4 y^{5} d y = \frac{2}{3}

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

= c (\frac{1}{3} - 1 + \frac{6}{5} - \frac{2}{3} + \frac{1}{7})

Simplify

c (\frac{1}{3} - 1 + \frac{6}{5} - \frac{2}{3} + \frac{1}{7}) : \frac{1}{105} c

= \frac{1}{105} c

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What is the integral from 0 to 1 of cy^2(1-y)^4 ?
The integral from 0 to 1 of cy^2(1-y)^4 is 1/105 c