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

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

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

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

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} x^{2} d y + \int_{- 1}^{1} y^{2} d y

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

\int_{- 1}^{1} y^{2} d y = \frac{2}{3}

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

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

\int_{1}^{e} (x + \frac{1}{x}) d x

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