On-line Math 21

5.2  The Fundamental Theorem of Calculus

Proposition 1 [MVT, integral form] If f(x) is continuous on [a,b] , then there is a c Î [a,b] so that

f(c) = ó õ b a  f(x)dx b-a .

Remark 2 If you are used to thinking about the definite integral as an antiderivative, then this is really just the old MVT applied to the antiderivative of f , but we're not there, just yet. This is a slightly different statement from the regular MVT, but is even easier to prove. We are assuming that any continuous function is integrable, which we proved earlier.

Proof:

Since f is continuous on [a,b] , it has a maximum value M and minimum value m on that interval,

m £ f(x) £ M, for all x Î [a,b].

But then, by the properties of integration,

m(b-a) £ ó õ b a  f(x)dx £ M(b-a),

or

m £ ó õ b a  f(x)dx b-a £ M.

However, any value between the maximum and minimum values of f on [a,b] is a value of f on [a,b] , by the Intermediate-Value Theorem, so there is a c Î [a,b] for which

f(c) = ó õ b a  f(x)dx b-a ,

which is what we needed to show.

Copyright (c) 2000 by David L. Johnson.