What is the integral from 0 to 2 of (6x-9)(4x^2+2) ?

The integral from 0 to 2 of (6x-9)(4x^2+2) is -12

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integral from 0 to 2 of (6x-9)(4x^2+2)

\int_{0}^{2} (6 x - 9) (4 x^{2} + 2) d x

- 12

Solution steps

\int_{0}^{2} (6 x - 9) (4 x^{2} + 2) d x

Expand

(6 x - 9) (4 x^{2} + 2) : 24 x^{3} + 12 x - 36 x^{2} - 18

= \int_{0}^{2} 24 x^{3} + 12 x - 36 x^{2} - 18 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} 24 x^{3} d x + \int_{0}^{2} 12 x d x - \int_{0}^{2} 36 x^{2} d x - \int_{0}^{2} 18 d x

\int_{0}^{2} 24 x^{3} d x = 96

\int_{0}^{2} 12 x d x = 24

\int_{0}^{2} 36 x^{2} d x = 96

\int_{0}^{2} 18 d x = 36

= 96 + 24 - 96 - 36

Simplify

= - 12

What is the integral from 0 to 2 of (6x-9)(4x^2+2) ?
The integral from 0 to 2 of (6x-9)(4x^2+2) is -12