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

The integral from 0 to 3 of (x^2-2xy+y^2) is 3x^2+9-9x

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

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

3 x^{2} + 9 - 9 x

Solution steps

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

\int_{0}^{3} x^{2} d y = 3 x^{2}

\int_{0}^{3} y^{2} d y = 9

\int_{0}^{3} 2 x y d y = 9 x

= 3 x^{2} + 9 - 9 x

\int_{1}^{2} (x - 1) - (x - 1)^{4} d x

\int_{0}^{1} x^{2} arccos (x) d x

\int_{0}^{1} x ln (x^{2} + 1) d x

\int_{\frac{1}{2}}^{1} (2 x - 1)^{2019} d x

\int_{0}^{1} 6 x x^{2} + 1 d x

What is the integral from 0 to 3 of (x^2-2xy+y^2) ?
The integral from 0 to 3 of (x^2-2xy+y^2) is 3x^2+9-9x