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

The integral from 0 to 4 of (2x^2-3xy+y^2) is 4y^2+128/3-24y

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integral from 0 to 4 of (2x^2-3xy+y^2)

\int_{0}^{4} (2 x^{2} - 3 x y + y^{2}) d x

4 y^{2} + \frac{128}{3} - 24 y

Solution steps

\int_{0}^{4} 2 x^{2} - 3 x y + 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} y^{2} d x + \int_{0}^{4} 2 x^{2} d x - \int_{0}^{4} 3 y x d x

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

\int_{0}^{4} 2 x^{2} d x = \frac{128}{3}

\int_{0}^{4} 3 y x d x = 24 y

= 4 y^{2} + \frac{128}{3} - 24 y

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What is the integral from 0 to 4 of (2x^2-3xy+y^2) ?
The integral from 0 to 4 of (2x^2-3xy+y^2) is 4y^2+128/3-24y