What is the integral from 2 to 5 of (6x^2+4y-3y^2) ?

The integral from 2 to 5 of (6x^2+4y-3y^2) is 18x^2-75

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integral from 2 to 5 of (6x^2+4y-3y^2)

\int_{2}^{5} (6 x^{2} + 4 y - 3 y^{2}) d y

18 x^{2} - 75

Solution steps

\int_{2}^{5} 6 x^{2} + 4 y - 3 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_{2}^{5} 3 y^{2} d y + \int_{2}^{5} 4 y d y + \int_{2}^{5} 6 x^{2} d y

\int_{2}^{5} 3 y^{2} d y = 117

\int_{2}^{5} 4 y d y = 42

\int_{2}^{5} 6 x^{2} d y = 18 x^{2}

= - 117 + 42 + 18 x^{2}

Simplify

= 18 x^{2} - 75

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What is the integral from 2 to 5 of (6x^2+4y-3y^2) ?
The integral from 2 to 5 of (6x^2+4y-3y^2) is 18x^2-75