What is the integral from 0 to 1 of 6/7 (x^2+(xy)/2) ?

The integral from 0 to 1 of 6/7 (x^2+(xy)/2) is (3y)/(14)+2/7

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integral from 0 to 1 of 6/7 (x^2+(xy)/2)

\int_{0}^{1} \frac{6}{7} (x^{2} + \frac{x y}{2}) d x

\frac{3 y}{14} + \frac{2}{7}

Solution steps

\int_{0}^{1} \frac{6}{7} (x^{2} + \frac{x y}{2}) d x

Take the constant out:

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

= \frac{6}{7} \cdot \int_{0}^{1} \frac{y x}{2} + 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{6}{7} (\int_{0}^{1} \frac{y x}{2} d x + \int_{0}^{1} x^{2} d x)

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

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

= \frac{6}{7} (\frac{y}{4} + \frac{1}{3})

Simplify

\frac{6}{7} (\frac{y}{4} + \frac{1}{3}) : \frac{3 y}{14} + \frac{2}{7}

= \frac{3 y}{14} + \frac{2}{7}

What is the integral from 0 to 1 of 6/7 (x^2+(xy)/2) ?
The integral from 0 to 1 of 6/7 (x^2+(xy)/2) is (3y)/(14)+2/7