What is the integral from 0 to 2 of (2x-9)(4x^2+3) ?

The integral from 0 to 2 of (2x-9)(4x^2+3) is -106

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integral from 0 to 2 of (2x-9)(4x^2+3)

\int_{0}^{2} (2 x - 9) (4 x^{2} + 3) d x

- 106

Solution steps

\int_{0}^{2} (2 x - 9) (4 x^{2} + 3) d x

Expand

(2 x - 9) (4 x^{2} + 3) : 8 x^{3} + 6 x - 36 x^{2} - 27

= \int_{0}^{2} 8 x^{3} + 6 x - 36 x^{2} - 27 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} 8 x^{3} d x + \int_{0}^{2} 6 x d x - \int_{0}^{2} 36 x^{2} d x - \int_{0}^{2} 27 d x

\int_{0}^{2} 8 x^{3} d x = 32

\int_{0}^{2} 6 x d x = 12

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

\int_{0}^{2} 27 d x = 54

= 32 + 12 - 96 - 54

Simplify

= - 106

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What is the integral from 0 to 2 of (2x-9)(4x^2+3) ?
The integral from 0 to 2 of (2x-9)(4x^2+3) is -106