Upper and lower bounds of integral

  • #1
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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!
 

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  • #2
Math_QED
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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.
 
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  • #3
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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.
 
  • #4
Math_QED
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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.
I don't see absolute values in the formula itself though.
 
  • #5
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Consider the function

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

Then

$$\int^1_0 x(1-x) dx = \frac{x^2}{2}-\frac{x^3}{3} = \frac{1}{6} > \frac{f(1)+f(0)}{2} = 0$$
 
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  • #6
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ok, what about the first part of the formula?
 
  • #7
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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.
In order to calculate that area ,
one needs to divide the parts.
@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##
ddddd28 said:
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.
 
  • #8
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For the first equation. Consider the function

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

$$2 \pi \sqrt {1 \times 1} >\int^{2\pi}_0\cos(x ) dx=0$$
 
  • #9
mathman
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Is it always true that:
View attachment 105614

View attachment 105615
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!

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]
 
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