Question

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

Answer 1

The answer is `=-3/4`

Explanation:

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`