How do you evaluate the definite integral $int (1/2x-1) dx$ from $[0, 3]$ ?

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Answer 1 · 2016-11-02T07:35:21

The answer is $=-3/4$

Use $intx^ndx=x^(n+1)/(n+1)+C$
$int_0^3(x/2-1)dx=int_0^3x/2dx-int_0^3 1*dx=(x^2/4-x)_0^3$
$=(9/4-3)-0=-3/4$

How do you evaluate the definite integral int (1/2x-1) dxint (1/2x-1) dx from [0, 3][0, 3]?