What is the integral from 0 to 6 of (9x^2+x-5) ?

The integral from 0 to 6 of (9x^2+x-5) is 636

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integral from 0 to 6 of (9x^2+x-5)

\int_{0}^{6} (9 x^{2} + x - 5) d x

636

Solution steps

\int_{0}^{6} 9 x^{2} + x - 5 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}^{6} 9 x^{2} d x + \int_{0}^{6} x d x - \int_{0}^{6} 5 d x

\int_{0}^{6} 9 x^{2} d x = 648

\int_{0}^{6} x d x = 18

\int_{0}^{6} 5 d x = 30

= 648 + 18 - 30

Simplify

= 636

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What is the integral from 0 to 6 of (9x^2+x-5) ?
The integral from 0 to 6 of (9x^2+x-5) is 636