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

The integral from 0 to 2 of (x^2+y^2) is 2y^2+8/3

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

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

2 y^{2} + \frac{8}{3}

Solution steps

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

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

= 2 y^{2} + \frac{8}{3}

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\int_{- π}^{π} x e^{- in x} d x

What is the integral from 0 to 2 of (x^2+y^2) ?
The integral from 0 to 2 of (x^2+y^2) is 2y^2+8/3