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

The integral from-2 to 2 of (x^2-y^2) is 4x^2-16/3

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

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

4 x^{2} - \frac{16}{3}

Solution steps

\int_{- 2}^{2} 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_{- 2}^{2} x^{2} d y - \int_{- 2}^{2} y^{2} d y

\int_{- 2}^{2} x^{2} d y = 4 x^{2}

\int_{- 2}^{2} y^{2} d y = \frac{16}{3}

= 4 x^{2} - \frac{16}{3}

What is the integral from-2 to 2 of (x^2-y^2) ?
The integral from-2 to 2 of (x^2-y^2) is 4x^2-16/3