What is the integral from 0 to 2 of (2x-6)(4x^2+7) ?

The integral from 0 to 2 of (2x-6)(4x^2+7) is -88

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integral from 0 to 2 of (2x-6)(4x^2+7)

\int_{0}^{2} (2 x - 6) (4 x^{2} + 7) d x

- 88

Solution steps

\int_{0}^{2} (2 x - 6) (4 x^{2} + 7) d x

Expand

(2 x - 6) (4 x^{2} + 7) : 8 x^{3} + 14 x - 24 x^{2} - 42

= \int_{0}^{2} 8 x^{3} + 14 x - 24 x^{2} - 42 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} 14 x d x - \int_{0}^{2} 24 x^{2} d x - \int_{0}^{2} 42 d x

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

\int_{0}^{2} 14 x d x = 28

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

\int_{0}^{2} 42 d x = 84

= 32 + 28 - 64 - 84

Simplify

= - 88

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What is the integral from 0 to 2 of (2x-6)(4x^2+7) ?
The integral from 0 to 2 of (2x-6)(4x^2+7) is -88