What is the integral from 0 to 2 of 16-x^2-2y^2 ?

The integral from 0 to 2 of 16-x^2-2y^2 is 32-8/3-4y^2

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integral from 0 to 2 of 16-x^2-2y^2

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

32 - \frac{8}{3} - 4 y^{2}

Solution steps

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

\int_{0}^{2} 16 d x = 32

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

\int_{0}^{2} 2 y^{2} d x = 4 y^{2}

= 32 - \frac{8}{3} - 4 y^{2}

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What is the integral from 0 to 2 of 16-x^2-2y^2 ?
The integral from 0 to 2 of 16-x^2-2y^2 is 32-8/3-4y^2