# How do you find int_-3^2 1/2(x+3)(x-2)?

Mar 19, 2016

${\int}_{- 3}^{2} \frac{1}{2} \left(x + 3\right) \left(x - 2\right) \mathrm{dx} = - \frac{125}{12}$

#### Explanation:

We will use the following:

• ${\int}_{a}^{b} \left(f + g\right) = {\int}_{a}^{b} f + {\int}_{a}^{b} g$ for integrable functions $f$ and $g$
• ${\int}_{a}^{b} c f = c {\int}_{a}^{b} f$ for an integrable function $f$ and constant $c$
• ${\int}_{a}^{b} {x}^{n} \mathrm{dx} = \frac{{b}^{n + 1} - {a}^{n + 1}}{n + 1}$ for $n \ne - 1$

With those, we have:

${\int}_{- 3}^{2} \frac{1}{2} \left(x + 3\right) \left(x - 2\right) \mathrm{dx} = \frac{1}{2} {\int}_{- 3}^{2} \left(x + 3\right) \left(x - 2\right) \mathrm{dx}$

$= \frac{1}{2} {\int}_{- 3}^{2} \left({x}^{2} + x - 6\right) \mathrm{dx}$

$= \frac{1}{2} \left({\int}_{- 3}^{2} {x}^{2} \mathrm{dx} + {\int}_{- 3}^{2} {x}^{1} \mathrm{dx} - 6 {\int}_{- 3}^{2} {x}^{0} \mathrm{dx}\right)$

$= \frac{1}{2} \left(\frac{{2}^{3} - {\left(- 3\right)}^{3}}{3} + \frac{{2}^{2} - {\left(- 3\right)}^{2}}{2} - \frac{6 \left(2 - \left(- 3\right)\right)}{1}\right)$

$= \frac{1}{2} \left(\frac{35}{3} - \frac{5}{2} - 30\right)$

$= - \frac{125}{12}$