What is the integral from 0 to 2 of (6x-3)(4x^2+4) ?

The integral from 0 to 2 of (6x-3)(4x^2+4) is 88

Popular >

integral from 0 to 2 of (6x-3)(4x^2+4)

\int_{0}^{2} (6 x - 3) (4 x^{2} + 4) d x

88

Solution steps

\int_{0}^{2} (6 x - 3) (4 x^{2} + 4) d x

Expand

(6 x - 3) (4 x^{2} + 4) : 24 x^{3} + 24 x - 12 x^{2} - 12

= \int_{0}^{2} 24 x^{3} + 24 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} 24 x^{3} d x + \int_{0}^{2} 24 x d x - \int_{0}^{2} 12 x^{2} d x - \int_{0}^{2} 12 d x

\int_{0}^{2} 24 x^{3} d x = 96

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

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

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

= 96 + 48 - 32 - 24

Simplify

= 88

\int_{1}^{2} \frac{5}{x ^{3}} d x

\int_{1}^{8} x e^{x^{2}} d x

\int_{1}^{\infty} \frac{2}{x ^{2} + 3} d x

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

\int_{0}^{1} e^{x} + x^{e} d x

What is the integral from 0 to 2 of (6x-3)(4x^2+4) ?
The integral from 0 to 2 of (6x-3)(4x^2+4) is 88