Prove Integral \int_{0}^{x}\frac{\sin(t)}{t+1}dt > 0 Without Calculus

  • Thread starter Thread starter Buri
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The integral \(\int_{0}^{x}\frac{\sin(t)}{t+1}dt\) is proven to be greater than zero for \(x > 0\) without using calculus or the Fundamental Theorem of Calculus. The solution involves analyzing the oscillatory behavior of the function \(\frac{\sin(t)}{t+1}\) and demonstrating that the integral from \([0,\pi]\) is greater in absolute value than the integral from \([\pi,2\pi]\). This pairing of consecutive intervals effectively establishes that the integral remains positive as \(x\) increases.

PREREQUISITES
  • Understanding of integral calculus concepts
  • Familiarity with oscillatory functions
  • Knowledge of properties of sine functions
  • Ability to analyze functions over specific intervals
NEXT STEPS
  • Study the properties of oscillatory functions in calculus
  • Learn about integral comparison tests
  • Explore techniques for proving inequalities involving integrals
  • Investigate the behavior of \(\sin(t)\) and its impact on integrals
USEFUL FOR

Students studying calculus, particularly those interested in integral inequalities, and educators looking for non-calculus methods to demonstrate integral properties.

Buri
Messages
271
Reaction score
0

Homework Statement



Prove that \int_{0}^{x}\frac{\sin(t)}{t+1}dt > 0

Homework Equations



It asks me to do this without actually calculating it and I can't use the Fundamental Theorem of Calculus.

The Attempt at a Solution



Okay, I see why this should be true as the function oscillates (with smaller bumps over time) around the x-axis so only the first bump really counts. I've tried considering small intervals where I know the function has certain values, but I can't seem to get anywhere with that idea either. Any help?
 
Last edited:
Physics news on Phys.org
Even a hint would be greatly appreciated! Thanks!
 
Ughhh I figured it out! I just showed that the integral from [0,pi] is larger in absolute value than the integral from [pi,2pi]. Continuing in this fashion (i.e. pairing up consecutive intervals) proves the assertion.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
Replies
9
Views
2K
Replies
6
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
4K