jaysscholar Posted March 1, 2005 Report 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!!!
Aki Posted March 1, 2005 Report 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
Bo Posted March 1, 2005 Report Posted March 1, 2005 the tricky part is to show that such a point c always exists :eek: (if f is continuous)
Qfwfq Posted March 1, 2005 Report 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.
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