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

The integral from 0 to 2 of (x^2+3xy+2y^2) is 2x^2+16/3+6x

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

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

2 x^{2} + \frac{16}{3} + 6 x

Solution steps

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

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

\int_{0}^{2} 2 y^{2} d y = \frac{16}{3}

\int_{0}^{2} 3 x y d y = 6 x

= 2 x^{2} + \frac{16}{3} + 6 x

What is the integral from 0 to 2 of (x^2+3xy+2y^2) ?
The integral from 0 to 2 of (x^2+3xy+2y^2) is 2x^2+16/3+6x