What is the integral from 0 to 2 of (2x-3)(8x^2+8) ?

The integral from 0 to 2 of (2x-3)(8x^2+8) is -16

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integral from 0 to 2 of (2x-3)(8x^2+8)

\int_{0}^{2} (2 x - 3) (8 x^{2} + 8) d x

- 16

Solution steps

\int_{0}^{2} (2 x - 3) (8 x^{2} + 8) d x

Expand

(2 x - 3) (8 x^{2} + 8) : 16 x^{3} + 16 x - 24 x^{2} - 24

= \int_{0}^{2} 16 x^{3} + 16 x - 24 x^{2} - 24 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} 16 x d x - \int_{0}^{2} 24 x^{2} d x - \int_{0}^{2} 24 d x

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

\int_{0}^{2} 16 x d x = 32

\int_{0}^{2} 24 x^{2} d x = 64

\int_{0}^{2} 24 d x = 48

= 64 + 32 - 64 - 48

Simplify

= - 16

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What is the integral from 0 to 2 of (2x-3)(8x^2+8) ?
The integral from 0 to 2 of (2x-3)(8x^2+8) is -16