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

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

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

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

y^{2} + \frac{1}{3}

Solution steps

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

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

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

= y^{2} + \frac{1}{3}

\int_{1}^{5} 3 x + 1 d x

\int_{0}^{\frac{15}{16}} 1 + 16 x d x

\int_{1}^{\infty} x^{1} d x

\int_{0}^{\frac{π}{3}} (1 + cos (x))^{2} d x

\int_{0}^{1} u^{2} d u

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