How Does Monotonicity Affect Integral Inequalities?

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SUMMARY

The discussion focuses on comparing the squared integral of a monotonically decreasing function \( f(x) \) over the interval [2, x] with the integral of the square of the function itself. Specifically, it examines the relationship between \([∫f(t)dt]^2\) and \(∫[f(t)]^2dt\) for functions such as \( f(x) = \frac{1}{x} \). The conclusion drawn suggests that as \( x \) approaches infinity, \([∫f(t)dt]^2\) is likely to be greater than \(∫[f(t)]^2dt\), although a formal proof is sought.

PREREQUISITES
  • Understanding of monotonic functions, specifically monotonically decreasing functions.
  • Familiarity with integral calculus, particularly properties of definite integrals.
  • Knowledge of the Mean Value Theorem for integrals.
  • Basic experience with evaluating integrals of functions like \( f(x) = \frac{1}{x} \).
NEXT STEPS
  • Study the properties of monotonic functions in relation to integrals.
  • Learn about the Mean Value Theorem for integrals and its applications.
  • Explore the concept of integral inequalities, particularly in the context of monotonic functions.
  • Investigate formal proofs related to integral comparisons, focusing on cases with decreasing functions.
USEFUL FOR

Mathematics students, educators, and researchers interested in integral calculus, particularly those studying properties of monotonic functions and integral inequalities.

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Homework Statement


suppose f(x) is monotonely decreasing and positive on [2,+∞),
please compare [∫f(t)dt]^2 and ∫[f(t)]^2dt,
here "∫ "means integrating on the interval [2,x]

Homework Equations


none


The Attempt at a Solution



Maybe the second mean value thereom of integral is helpful.
 
Last edited:
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Have you tried anything? In particular, have you selected some simple monotonically decreasing function, such as f(x)= \frac{1}{x} and calculated those two values?
 
In fact, yes!
But what I'm really eager to know is how to prove the conclusion.
Maybe when the x is large enough, [∫f(t)dt]^2 is larger.
 

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