How Does the Integral Mean Value Theorem Link to Fundamental Calculus Concepts?

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SUMMARY

The Integral Mean Value Theorem establishes a direct relationship between the ordinary Mean Value Theorem and the Fundamental Theorem of Calculus. Specifically, it can be expressed as \(\int_a^b G'(x)dx=(b-a)G'(\xi)\), demonstrating that the average rate of change of a function over an interval corresponds to the integral of its derivative. This discussion highlights the circularity in using the Mean Value Theorem to prove the Fundamental Theorem of Calculus, emphasizing the need for a clear understanding of these foundational concepts in calculus.

PREREQUISITES
  • Understanding of the Mean Value Theorem
  • Familiarity with the Fundamental Theorem of Calculus
  • Knowledge of integral calculus
  • Ability to interpret mathematical proofs
NEXT STEPS
  • Study the proof of the Mean Value Theorem in detail
  • Explore the Fundamental Theorem of Calculus and its applications
  • Learn about the implications of the Integral Mean Value Theorem
  • Investigate potential circular reasoning in calculus proofs
USEFUL FOR

Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of the relationships between fundamental calculus concepts.

daudaudaudau
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http://en.wikipedia.org/wiki/Mean_value_theorem#First_mean_value_theorem_for_integration"

Take a look at the Wikipedia proof. Now, wouldn't it be easier to prove it like this:
The ordinary mean value theorem says that
[tex] G(b)-G(a)=(b-a)G'(\xi)[/tex]

And the fundamental theorem of calculus says that
[tex] G(b)-G(a)=\int_a^b G'(x)dx[/tex]

So the conclusion is
[tex] \int_a^b G'(x)dx=(b-a)G'(\xi)[/tex]
 
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You proved the case where [tex]\phi = 1[/tex] identically. Notice the wikipedia proof covers a far broader case than you did.

Also, you probably used the mean value theorem to prove the fundamental theorem of calculus, so this is circular
 
I see.
 

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