What is the integral from 0 to 2 of (2x-9)(4x^2+4) ?

The integral from 0 to 2 of (2x-9)(4x^2+4) is -120

integral from 0 to 2 of (2x-9)(4x^2+4)

\int_{0}^{2} (2 x - 9) (4 x^{2} + 4) d x

- 120

Solution steps

\int_{0}^{2} (2 x - 9) (4 x^{2} + 4) d x

Expand

(2 x - 9) (4 x^{2} + 4) : 8 x^{3} + 8 x - 36 x^{2} - 36

= \int_{0}^{2} 8 x^{3} + 8 x - 36 x^{2} - 36 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} 36 x^{2} d x - \int_{0}^{2} 36 d x

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

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

\int_{0}^{2} 36 x^{2} d x = 96

\int_{0}^{2} 36 d x = 72

= 32 + 16 - 96 - 72

Simplify

= - 120

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

\int_{0}^{2} \frac{x}{1 + 6 x ^{2}} d x

\int_{0}^{\frac{π}{4}} (sec (x))^{3} d x

\int_{1}^{2} 3 x^{3} - 9 x^{2} + 6 x d x

\int_{0}^{1} x (1 - x^{2})^{9} d x

What is the integral from 0 to 2 of (2x-9)(4x^2+4) ?
The integral from 0 to 2 of (2x-9)(4x^2+4) is -120