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

The integral from 0 to 4 of (4x^2+xy-2y^2) is -8y^2+256/3+y*8

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

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

- 8 y^{2} + \frac{256}{3} + y \cdot 8

Solution steps

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

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

\int_{0}^{4} 4 x^{2} d x = \frac{256}{3}

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

= - 8 y^{2} + \frac{256}{3} + y \cdot 8

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What is the integral from 0 to 4 of (4x^2+xy-2y^2) ?
The integral from 0 to 4 of (4x^2+xy-2y^2) is -8y^2+256/3+y*8