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Find the integral of sqrt(x^2-2x+1) dx from 0 to 1. (Answer: 1/2)
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factor the integrand
sqrt(x^2-2x+1)= sqrt(x-1)^2 so that leaves (x-1)dx from 0 to 1
And then?
= [(x^2)/2] from 0 to 1 = 1/2
do you get it?
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So the integral of (x-1) dx is not x^2/2-x?
sorry im drunk i will tell you now the correct one :D
its absolute(x-1) because we too sqrt off so from 0 to 1 absoulute(x-1) = -x+1 so the integral becomes [x-(x^2)/2] from 0 to 1 = 1/2 :D
took*
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