What is the integral from 0 to 1 of 6y(1-y) ?

The integral from 0 to 1 of 6y(1-y) is 1

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integral from 0 to 1 of 6y(1-y)

\int_{0}^{1} 6 y (1 - y) d y

1

Solution steps

\int_{0}^{1} 6 y (1 - y) d y

Take the constant out:

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

= 6 \cdot \int_{0}^{1} y (1 - y) d y

Expand

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

= 6 \cdot \int_{0}^{1} y - y^{2} 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

= 6 (\int_{0}^{1} y d y - \int_{0}^{1} y^{2} d y)

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

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

= 6 (\frac{1}{2} - \frac{1}{3})

Simplify

6 (\frac{1}{2} - \frac{1}{3}) : 1

= 1

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What is the integral from 0 to 1 of 6y(1-y) ?
The integral from 0 to 1 of 6y(1-y) is 1