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

The integral from 0 to 2 of (2x-3)(4x^2+4) is -8

Popular >

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

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

- 8

Solution steps

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

Expand

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

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

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

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

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

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

= 32 + 16 - 32 - 24

Simplify

= - 8

\int_{0}^{2} \frac{x ^{3}}{16 - x ^{2}} d x

\int_{- 4}^{4} 2 x d x

\int_{1}^{2} 2 x (x^{2} + 6) d x

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

\int_{1}^{2} 1 + 4 x^{2} d x

What is the integral from 0 to 2 of (2x-3)(4x^2+4) ?
The integral from 0 to 2 of (2x-3)(4x^2+4) is -8