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How do you find the integral from 0 to 3 of #(1/2 x - 1) dx#?

1 Answer

Apr 13, 2015

#int_0^3(1/2x-1)dx = int_0^3(1/4x^2-x)dx#

We have :

#1/2[(1/2x^2-2x)]_0^3#

Therefore :

#1/2(9/2-6)#

#=-3/4#

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