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

The integral from 0 to 2 of (6x-6)(8x^2+2) is 64

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

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

64

Solution steps

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

Expand

(6 x - 6) (8 x^{2} + 2) : 48 x^{3} + 12 x - 48 x^{2} - 12

= \int_{0}^{2} 48 x^{3} + 12 x - 48 x^{2} - 12 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} 48 x^{3} d x + \int_{0}^{2} 12 x d x - \int_{0}^{2} 48 x^{2} d x - \int_{0}^{2} 12 d x

\int_{0}^{2} 48 x^{3} d x = 192

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

\int_{0}^{2} 48 x^{2} d x = 128

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

= 192 + 24 - 128 - 24

Simplify

= 64

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What is the integral from 0 to 2 of (6x-6)(8x^2+2) ?
The integral from 0 to 2 of (6x-6)(8x^2+2) is 64