What is the integral from 0 to 2 of (2x-9)(8x^2+6) ?

The integral from 0 to 2 of (2x-9)(8x^2+6) is -212

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integral from 0 to 2 of (2x-9)(8x^2+6)

\int_{0}^{2} (2 x - 9) (8 x^{2} + 6) d x

- 212

Solution steps

\int_{0}^{2} (2 x - 9) (8 x^{2} + 6) d x

Expand

(2 x - 9) (8 x^{2} + 6) : 16 x^{3} + 12 x - 72 x^{2} - 54

= \int_{0}^{2} 16 x^{3} + 12 x - 72 x^{2} - 54 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} 16 x^{3} d x + \int_{0}^{2} 12 x d x - \int_{0}^{2} 72 x^{2} d x - \int_{0}^{2} 54 d x

\int_{0}^{2} 16 x^{3} d x = 64

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

\int_{0}^{2} 72 x^{2} d x = 192

\int_{0}^{2} 54 d x = 108

= 64 + 24 - 192 - 108

Simplify

= - 212

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What is the integral from 0 to 2 of (2x-9)(8x^2+6) ?
The integral from 0 to 2 of (2x-9)(8x^2+6) is -212