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

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

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

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

\frac{6}{7} (x + 2 x^{2})

Solution steps

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

Take the constant out:

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

= \frac{6}{7} \cdot \int_{0}^{2} \frac{x y}{2} + x^{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

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

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

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

= \frac{6}{7} (x + 2 x^{2})

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