What is the integral from 0 to 3 of (10x-6x^2+9) ?

The integral from 0 to 3 of (10x-6x^2+9) is 18

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integral from 0 to 3 of (10x-6x^2+9)

\int_{0}^{3} (10 x - 6 x^{2} + 9) d x

18

Solution steps

\int_{0}^{3} 10 x - 6 x^{2} + 9 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}^{3} 10 x d x - \int_{0}^{3} 6 x^{2} d x + \int_{0}^{3} 9 d x

\int_{0}^{3} 10 x d x = 45

\int_{0}^{3} 6 x^{2} d x = 54

\int_{0}^{3} 9 d x = 27

= 45 - 54 + 27

Simplify

= 18

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What is the integral from 0 to 3 of (10x-6x^2+9) ?
The integral from 0 to 3 of (10x-6x^2+9) is 18