What is the integral from 0 to 1 of 12/5 x(2-x-y) ?

The integral from 0 to 1 of 12/5 x(2-x-y) is (2(-3y+4))/5

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integral from 0 to 1 of 12/5 x(2-x-y)

\int_{0}^{1} \frac{12}{5} x (2 - x - y) d x

\frac{2 ( - 3 y + 4 )}{5}

Solution steps

\int_{0}^{1} \frac{12}{5} x (2 - x - y) d x

Take the constant out:

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

= \frac{12}{5} \cdot \int_{0}^{1} (2 - y - x) x d x

Expand

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

= \frac{12}{5} \cdot \int_{0}^{1} 2 x - y 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

= \frac{12}{5} (\int_{0}^{1} 2 x d x - \int_{0}^{1} y x d x - \int_{0}^{1} x^{2} d x)

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

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

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

= \frac{12}{5} (1 - \frac{y}{2} - \frac{1}{3})

Simplify

\frac{12}{5} (1 - \frac{y}{2} - \frac{1}{3}) : \frac{2 ( - 3 y + 4 )}{5}

= \frac{2 ( - 3 y + 4 )}{5}

What is the integral from 0 to 1 of 12/5 x(2-x-y) ?
The integral from 0 to 1 of 12/5 x(2-x-y) is (2(-3y+4))/5