What is the integral from 0 to 1 of 12xy(1-x) ?

The integral from 0 to 1 of 12xy(1-x) is 2y

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integral from 0 to 1 of 12xy(1-x)

\int_{0}^{1} 12 x y (1 - x) d x

2 y

Solution steps

\int_{0}^{1} 12 x y (1 - x) d x

Take the constant out:

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

= 12 y \cdot \int_{0}^{1} (1 - x) x d x

Expand

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

= 12 y \cdot \int_{0}^{1} x - x^{2} 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

= 12 y (\int_{0}^{1} x d x - \int_{0}^{1} x^{2} d x)

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

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

= 12 y (\frac{1}{2} - \frac{1}{3})

Simplify

12 y (\frac{1}{2} - \frac{1}{3}) : 2 y

= 2 y

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What is the integral from 0 to 1 of 12xy(1-x) ?
The integral from 0 to 1 of 12xy(1-x) is 2y