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

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Answer 1 · 2016-12-03T13:07:20

6

#int_1^3(2x-1)dx#

#=[(cancel(2)x^2)/cancel(2)-x]_1^3#

#=[x^2-x]^3-[x^2-x]_1#

#=3^2-3-cancel((1^2-1))^0#

#=6#