Question

How do you integrate `(1/2)x-1dx` for the interval (0,3)?

Answer 1

I found: `-3/4`

Explanation:

Given:
`int_0^3(x/2-1)dx=`
`[x^2/4-x]_0^3=` substituting the extrema:
`=(9/4-3)-0=(9-12)/4=-3/4`