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

The integral from-1 to 2 of (3x-2)(x+1) is 9/2

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

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

\frac{9}{2}

Decimal

4.5

Solution steps

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

Expand

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

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

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

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

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

= 9 + \frac{3}{2} - 6

Simplify

9 + \frac{3}{2} - 6 : \frac{9}{2}

= \frac{9}{2}

What is the integral from-1 to 2 of (3x-2)(x+1) ?
The integral from-1 to 2 of (3x-2)(x+1) is 9/2