jaysscholar Posted March 1, 2005 Report Share Posted March 1, 2005 If f is continuous on the closed interval [a,b], then there exists c such that a<c<b and the integral of f(x) dx from a to b=....? a) f©/(b-a) :eek: (f(:)-f(a))/(b-a) c) f(:eek:-f(a) d) f'©(b-a) e) f©(b-a) help!!! Link to comment Share on other sites More sharing options...
Aki Posted March 1, 2005 Report Share Posted March 1, 2005 Hey jay, welcome to the forums. The answer is e) f©(b-a)The integral is basicaly the area under the curve. So f© is your height, and (b-a) is your width. When you multiply those together, you get the area, or the integral.That's the Mean Value Theorem for integrals pgrmdave 1 Link to comment Share on other sites More sharing options...
Bo Posted March 1, 2005 Report Share Posted March 1, 2005 the tricky part is to show that such a point c always exists :eek: (if f is continuous) Link to comment Share on other sites More sharing options...
Qfwfq Posted March 1, 2005 Report Share Posted March 1, 2005 I believe this is called the Rolle theorem and it is related to the Lagrange theorem. A mean value can't be greater than the max (or sup) nor less than the min or inf. As the interval is closed and f continuous, it will have a finite max and min. Continuity then allows to apply the theorem that, for any value v between f(d) and f(e), there will be a point g, between d and e, for which f has the value v. Link to comment Share on other sites More sharing options...
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