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

The integral from 1 to 2 of (x-3)(x+6) is -67/6

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

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

- \frac{67}{6}

Decimal

- 11.16666 \dots

Solution steps

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

Expand

(x - 3) (x + 6) : x^{2} + 3 x - 18

= \int_{1}^{2} x^{2} + 3 x - 18 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} x^{2} d x + \int_{1}^{2} 3 x d x - \int_{1}^{2} 18 d x

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

\int_{1}^{2} 18 d x = 18

= \frac{7}{3} + \frac{9}{2} - 18

Simplify

\frac{7}{3} + \frac{9}{2} - 18 : - \frac{67}{6}

= - \frac{67}{6}

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What is the integral from 1 to 2 of (x-3)(x+6) ?
The integral from 1 to 2 of (x-3)(x+6) is -67/6