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Is this inequality true and provable?

by 3029298
Tags: inequality, integral, mathematics, proof
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Aug8-08, 02:06 PM
P: 57
1. The problem statement, all variables and given/known data
My question is whether the following inequality can be proven.

2. Relevant equations

3. The attempt at a solution
I tried to write down the inequality in the form of it's primitives, where [tex]G\left(x\right)[/tex] is the primitive of [tex]g\left(x\right)[/tex] and [tex]H\left(x\right)[/tex] is the primitive of [tex]h\left(x\right)[/tex]. The inequality then becomes:


But what next, or are there other means of getting a proof?
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Aug8-08, 02:59 PM
P: 1,104
Assuming [tex] a \leq b [/tex] and f is continuous on the interval [a,b], then

[tex] \left|\int_a^bf\left(x\right)dx \right| \leq \int_a^b\left|f(x)\right|dx[/tex]

which follows from the fact that [tex] f(x) \leq \left|f(x)\right| [/tex] and [tex] -f(x) \leq \left|f(x)\right| [/tex] and that

If f,g are both continuous on the interval [a,b] and [tex] f(x) \leq g(x) [/tex] for all x in the interval. Then

[tex] \int_a^b f(x)dx \leq \int_a^b g(x)dx [/tex]

Rearranging and using the first inequality should give you the desired inequality.
Aug8-08, 03:08 PM
P: 57
Oh, I see it now, it is indeed not that difficult.


If we rearrange:


Substituting [tex]f\left(x\right)=g\left(x\right)-h\left(x\right)[/tex] and using the first formula of snipez90, we get the proof.


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