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

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

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

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

\frac{y}{6} + \frac{1}{3}

Solution steps

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

\int_{0}^{1} \frac{y x}{3} d x = \frac{y}{6}

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

= \frac{y}{6} + \frac{1}{3}

\int_{2}^{\infty} \frac{x}{x ^{2} - 3} d x

\int_{0}^{3} x^{2} cos (\frac{nπ x}{3}) d x

\int_{0}^{2} 16 x - x^{5} d x

\int_{1}^{2} \frac{3}{7} x^{2} x^{2} d x

\int_{2}^{4} 3 + x^{2} d x

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