Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Upper and lower bounds of integral

Tags:
  1. Sep 8, 2016 #1
    Is it always true that:
    upload_2016-9-8_20-37-30.png

    upload_2016-9-8_20-38-26.png
    Noticing that it works for some functions, I wanted to ensure it is true for all of them( at least polynomical), but since I am still in highschool, and I don't have deep understanding in calculus( yet), the question is forwarded to you. proof please!
     

    Attached Files:

    Last edited by a moderator: Sep 8, 2016
  2. jcsd
  3. Sep 8, 2016 #2

    Math_QED

    User Avatar
    Homework Helper

    The first equation is obviously false. Consider: ##f: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto x.##

    Then, ##\int\limits_{-1}^1 x dx = 0##, while ##(b-a) \sqrt{f(a) f(b)} = 2 \sqrt{-1} \quad ##is not defined when we only allow real numbers.
     
    Last edited: Sep 8, 2016
  4. Sep 8, 2016 #3
    In order to calculate that area , one needs to divide the parts. Let's assume also that only the absolute values are taken, and then , there will not be any problems with the root output. Thus, the formula is still valid.
     
  5. Sep 8, 2016 #4

    Math_QED

    User Avatar
    Homework Helper

    I don't see absolute values in the formula itself though.
     
  6. Sep 8, 2016 #5
    Consider the function

    $$f(x) = x(1-x)$$

    Then

    $$\int^1_0 x(1-x) dx = \frac{x^2}{2}-\frac{x^3}{3} = \frac{1}{6} > \frac{f(1)+f(0)}{2} = 0$$
     
  7. Sep 8, 2016 #6
    ok, what about the first part of the formula?
     
  8. Sep 8, 2016 #7

    Mark44

    Staff: Mentor

    @ddddd28, you posted an inequality that included a definite integral, which can represent an area, but doesn't have to.
    The inequality was ##(b - a)\sqrt{f(a)f(b)} \le \int_a^b f(x) dx##
     
  9. Sep 8, 2016 #8
    For the first equation. Consider the function

    $$f (x) = \cos(x), a= 0 , b= 2\pi$$

    $$2 \pi \sqrt {1 \times 1} >\int^{2\pi}_0\cos(x ) dx=0$$
     
  10. Sep 8, 2016 #9

    mathman

    User Avatar
    Science Advisor
    Gold Member


    Both false - simple counterexamples:
    [itex](1) f(x)=x^2,\ a=-1,\ b=1,\ (2) f(x)=1-x^2,\ a=-1,\ b=1[/itex]
     
    Last edited by a moderator: Sep 8, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Upper and lower bounds of integral
  1. Upper bound (Replies: 18)

Loading...