What is the integral from-1 to 2 of (3x-5)(x+2) ?

The integral from-1 to 2 of (3x-5)(x+2) is -39/2

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integral from-1 to 2 of (3x-5)(x+2)

\int_{- 1}^{2} (3 x - 5) (x + 2) d x

- \frac{39}{2}

Decimal

- 19.5

Solution steps

\int_{- 1}^{2} (3 x - 5) (x + 2) d x

Expand

(3 x - 5) (x + 2) : 3 x^{2} + x - 10

= \int_{- 1}^{2} 3 x^{2} + x - 10 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} 10 d x

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

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

\int_{- 1}^{2} 10 d x = 30

= 9 + \frac{3}{2} - 30

Simplify

9 + \frac{3}{2} - 30 : - \frac{39}{2}

= - \frac{39}{2}

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What is the integral from-1 to 2 of (3x-5)(x+2) ?
The integral from-1 to 2 of (3x-5)(x+2) is -39/2