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

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

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

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

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