# I Upper and lower bounds of integral

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1. Sep 8, 2016

### ddddd28

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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2. Sep 8, 2016

### Math_QED

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
3. Sep 8, 2016

### ddddd28

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. Sep 8, 2016

### Math_QED

I don't see absolute values in the formula itself though.

5. Sep 8, 2016

### zaidalyafey

Consider the function

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

Then

$$\int^1_0 x(1-x) dx = \frac{x^2}{2}-\frac{x^3}{3} = \frac{1}{6} > \frac{f(1)+f(0)}{2} = 0$$

6. Sep 8, 2016

### ddddd28

ok, what about the first part of the formula?

7. Sep 8, 2016

### 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 $(b - a)\sqrt{f(a)f(b)} \le \int_a^b f(x) dx$

8. Sep 8, 2016

### zaidalyafey

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. Sep 8, 2016

### mathman

Both false - simple counterexamples:
$(1) f(x)=x^2,\ a=-1,\ b=1,\ (2) f(x)=1-x^2,\ a=-1,\ b=1$

Last edited by a moderator: Sep 8, 2016