What is the integral from 0 to 1 of 6/5 (x+y^2) ?

The integral from 0 to 1 of 6/5 (x+y^2) is 6/5 x+2/5

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integral from 0 to 1 of 6/5 (x+y^2)

\int_{0}^{1} \frac{6}{5} (x + y^{2}) d y

\frac{6}{5} x + \frac{2}{5}

Solution steps

\int_{0}^{1} \frac{6}{5} (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}{5} \cdot \int_{0}^{1} x + 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

= \frac{6}{5} (\int_{0}^{1} x d y + \int_{0}^{1} y^{2} d y)

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

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

= \frac{6}{5} (x + \frac{1}{3})

Simplify

\frac{6}{5} (x + \frac{1}{3}) : \frac{6}{5} x + \frac{2}{5}

= \frac{6}{5} x + \frac{2}{5}

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What is the integral from 0 to 1 of 6/5 (x+y^2) ?
The integral from 0 to 1 of 6/5 (x+y^2) is 6/5 x+2/5