Answer to Question #45740 in Integral Calculus for victor
Determine the following integral
∫x−3dx
1
2014-09-05T14:17:05-0400
∫(x−3)dx=∫(x−3)d(x-3)=|substitution t=x-3|=∫tdt=(t^2)/2+C=(x-3)^2/2+C, where C is an arbitrary real constant
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