What is the integral from-1 to 2 of (4x^2-1)(6x+3) ?

The integral from-1 to 2 of (4x^2-1)(6x+3) is 108

Popular >

integral from-1 to 2 of (4x^2-1)(6x+3)

\int_{- 1}^{2} (4 x^{2} - 1) (6 x + 3) d x

108

Solution steps

\int_{- 1}^{2} (4 x^{2} - 1) (6 x + 3) d x

Expand

(4 x^{2} - 1) (6 x + 3) : 24 x^{3} + 12 x^{2} - 6 x - 3

= \int_{- 1}^{2} 24 x^{3} + 12 x^{2} - 6 x - 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_{- 1}^{2} 24 x^{3} d x + \int_{- 1}^{2} 12 x^{2} d x - \int_{- 1}^{2} 6 x d x - \int_{- 1}^{2} 3 d x

\int_{- 1}^{2} 24 x^{3} d x = 90

\int_{- 1}^{2} 12 x^{2} d x = 36

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

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

= 90 + 36 - 9 - 9

Simplify

= 108

\int_{0}^{x} arctan (t) d t

\int_{0}^{π} 0 d x

\int_{0}^{\frac{π}{4}} 4 sec^{2} (x) d x

\int_{0}^{\frac{π}{2}} sin^{3} (θ) d θ

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

What is the integral from-1 to 2 of (4x^2-1)(6x+3) ?
The integral from-1 to 2 of (4x^2-1)(6x+3) is 108