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

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

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

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

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

$=6$