What is the integral from 0 to 2 of (2x-3)(4x^2+1) ?

The integral from 0 to 2 of (2x-3)(4x^2+1) is -2

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integral from 0 to 2 of (2x-3)(4x^2+1)

\int_{0}^{2} (2 x - 3) (4 x^{2} + 1) d x

- 2

Solution steps

\int_{0}^{2} (2 x - 3) (4 x^{2} + 1) d x

Expand

(2 x - 3) (4 x^{2} + 1) : 8 x^{3} + 2 x - 12 x^{2} - 3

= \int_{0}^{2} 8 x^{3} + 2 x - 12 x^{2} - 3 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} 2 x d x - \int_{0}^{2} 12 x^{2} d x - \int_{0}^{2} 3 d x

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

\int_{0}^{2} 2 x d x = 4

\int_{0}^{2} 12 x^{2} d x = 32

\int_{0}^{2} 3 d x = 6

= 32 + 4 - 32 - 6

Simplify

= - 2

f (x) = 4 x^{2} + 4 x - 2, at x = - 1

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What is the integral from 0 to 2 of (2x-3)(4x^2+1) ?
The integral from 0 to 2 of (2x-3)(4x^2+1) is -2