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

The integral from 0 to 2 of (2x-6)(4x^2+4) is -64

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

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

- 64

Solution steps

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

Expand

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

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

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

\int_{0}^{2} 8 x d x = 16

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

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

= 32 + 16 - 64 - 48

Simplify

= - 64

\int_{1}^{2} 3 x^{2} - 5 x + 7 d x

\int_{0}^{1} y (e - 1) d y

\int_{- 8}^{0} (5 + 64 - x^{2}) d x

\int_{0}^{1} 3 - 3 x^{2} d x

\int_{- 1}^{1} (8 - x) d x

What is the integral from 0 to 2 of (2x-6)(4x^2+4) ?
The integral from 0 to 2 of (2x-6)(4x^2+4) is -64