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I Upper and lower bounds of integral

  1. Sep 8, 2016 #1
    Is it always true that:

    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!

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


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    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


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

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


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


    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


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    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
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