# How do you find the integral from 0 to 3 of (1/2 x - 1) dx?

Apr 13, 2015

${\int}_{0}^{3} \left(\frac{1}{2} x - 1\right) \mathrm{dx} = {\int}_{0}^{3} \left(\frac{1}{4} {x}^{2} - x\right) \mathrm{dx}$

We have :

$\frac{1}{2} {\left[\left(\frac{1}{2} {x}^{2} - 2 x\right)\right]}_{0}^{3}$

Therefore :

$\frac{1}{2} \left(\frac{9}{2} - 6\right)$

$= - \frac{3}{4}$