What is the integral from 6 to 3 of (9x^2-4x+9) ?

The integral from 6 to 3 of (9x^2-4x+9) is -540

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integral from 6 to 3 of (9x^2-4x+9)

\int_{6}^{3} (9 x^{2} - 4 x + 9) d x

- 540

Solution steps

\int_{6}^{3} 9 x^{2} - 4 x + 9 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{3}^{6} 9 x^{2} - 4 x + 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_{3}^{6} 9 x^{2} d x - \int_{3}^{6} 4 x d x + \int_{3}^{6} 9 d x)

\int_{3}^{6} 9 x^{2} d x = 567

\int_{3}^{6} 4 x d x = 54

\int_{3}^{6} 9 d x = 27

= - (567 - 54 + 27)

Simplify

= - 540

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What is the integral from 6 to 3 of (9x^2-4x+9) ?
The integral from 6 to 3 of (9x^2-4x+9) is -540