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

The integral from 0 to 2 of (3x^2+9-9x) is 8

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

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

8

Solution steps

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

\int_{0}^{2} 3 x^{2} d x = 8

\int_{0}^{2} 9 d x = 18

\int_{0}^{2} 9 x d x = 18

= 8 + 18 - 18

Simplify

= 8

What is the integral from 0 to 2 of (3x^2+9-9x) ?
The integral from 0 to 2 of (3x^2+9-9x) is 8