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

The integral from 0 to 1 of 3y(1-y)^2 is 1/4

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

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

\frac{1}{4}

Decimal

0.25

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{0}^{1} y (1 - y)^{2} d y

Expand

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

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

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

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

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

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

= 3 (\frac{1}{2} - \frac{2}{3} + \frac{1}{4})

Simplify

3 (\frac{1}{2} - \frac{2}{3} + \frac{1}{4}) : \frac{1}{4}

= \frac{1}{4}

What is the integral from 0 to 1 of 3y(1-y)^2 ?
The integral from 0 to 1 of 3y(1-y)^2 is 1/4