What is the integral from 0 to 4 of (x^2-2y^2) ?

The integral from 0 to 4 of (x^2-2y^2) is 64/3-8y^2

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integral from 0 to 4 of (x^2-2y^2)

\int_{0}^{4} (x^{2} - 2 y^{2}) d x

\frac{64}{3} - 8 y^{2}

Solution steps

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

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 2 y^{2} d x = 8 y^{2}

= \frac{64}{3} - 8 y^{2}

\int_{1}^{5} (2 x + \frac{1}{x}) d x

\int_{0}^{14} \frac{x}{25 + 4 x} d x

\int_{0}^{1} e^{x} - e d x

\int_{0}^{2 π} - 4 sin^{2} (x) d x

\int_{0}^{π} (2 + cos (x)) d x

What is the integral from 0 to 4 of (x^2-2y^2) ?
The integral from 0 to 4 of (x^2-2y^2) is 64/3-8y^2