What is the integral from 0 to 1 of ((4x)/3+y^2) ?

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

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integral from 0 to 1 of ((4x)/3+y^2)

\int_{0}^{1} (\frac{4 x}{3} + y^{2}) d x

\frac{2}{3} + y^{2}

Solution steps

\int_{0}^{1} \frac{4 x}{3} + y^{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

= \int_{0}^{1} \frac{4 x}{3} d x + \int_{0}^{1} y^{2} d x

\int_{0}^{1} \frac{4 x}{3} d x = \frac{2}{3}

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

= \frac{2}{3} + y^{2}

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