Corollary 2 If f¢(x) = g¢(x) for all x Î [a,b] , then, for some constant C , f(x) = g(x)+C .
Proof. Set h(x) = f(x)-g(x). Note that h is differentiable on (a,b) and continuous on [a,b] . Then the previous result implies that h(x) is a constant, since h¢(x) = f¢(x)-g¢(x) = 0. Adding g(x) to both sides gives the result.
Copyright (c) 2000 by David L. Johnson.