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

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Answer 1 · 2015-08-18T19:45:40

I found: $-3/4$

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

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