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

The integral from 0 to 3 of 2y-y^2-(y^2-4y) is 9

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

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

9

Solution steps

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

Simplify

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

\int_{0}^{3} 2 y^{2} d y = 18

\int_{0}^{3} 6 y d y = 27

= - 18 + 27

Simplify

= 9

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What is the integral from 0 to 3 of 2y-y^2-(y^2-4y) ?
The integral from 0 to 3 of 2y-y^2-(y^2-4y) is 9