Is this inequality true?

  • Thread starter jenga42
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  • #1
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Hello,

I'm trying to figure out if this is true - any help would be greatly appreciated!

If a(x) < b(x) between the values x_i and x_f then is the following also true?...

int_{x_i}^{x_f} dx f(x) a(x) < int_{x_i}^{x_f} dx f(x) b(x)

where f(x)>0 for all x.

...I think it is true, but I'd like to be able to prove it!

Thanks,

Jenny
 

Answers and Replies

  • #2
1,101
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Your notation is a bit awkward. Are f(x)a(x) and f(x)b(x) the integrands?
 
  • #3
CompuChip
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If f is continuous (you will definitely need that, I think) and the interval is finite, then you could probably estimate it by its maximum and do a series of inequalities like
[tex]| \int_a^b f(x) p(x) dx | < | f(x) p(x) (b - a) | < \cdots[/tex]

Or you could try considering
[tex]\int_a^b f(x) (p(x) - q(x) ) [/tex]
and showing that it is positive (for q(x) < p(x) everywhere).
 
  • #4
morphism
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You don't need continuity or anything of the sort -- the integral is monotone. That is, if h and k are integrable on [a,b], and h(x)<=k(x) for all x in [a,b], then

[tex]\int_a^b h(x) dx \leq \int_a^b k(x) dx.[/tex]

Edit: Oh. Maybe that's what you were saying here:
Or you could try considering
[tex]\int_a^b f(x) (p(x) - q(x) ) [/tex]
and showing that it is positive (for q(x) < p(x) everywhere).
 
  • #5
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Hi,

Thanks for replying! I didn't know I could type the equations!

Just to clarify, I meant if a(x)<b(x), can I say

[tex]\int_{x_i}^{x_f} f(x) a(x) dx < \int_{x_i}^{x_f} f(x) b(x) dx[/tex]

where f(x)>0.

I'd like to know if there is a general theorem for this to be true? ...I think that (please correct me if I'm wrong!) I could only use

Or you could try considering
[tex]\int_a^b f(x) (p(x) - q(x) )[/tex]
and showing that it is positive (for q(x) < p(x) everywhere).

for a specific f(x).

What does "the integral is monotone" mean?

Thanks!
 
  • #6
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I've just had a thought, can I just say that as f(x)>0, then it cannot change the direction of the inequality, hence if a(x)<b(x) , then multiplying both sides by f(x), we find f(x)a(x)<f(x)b(x) and therefore

[tex]\int_{x_i}^{x_f} f(x)a(x) dx < \int_{x_i}^{x_f} f(x)b(x) dx [/tex]

????

This is assuming that if a(x)<b(x), then

[tex]\int_{x_i}^{x_f} a(x) dx < \int_{x_i}^{x_f} b(x) dx [/tex]

But I don't understand why this second equation is true - I've tried looking up monotone functions, but haven't found anything useful!

Thanks
 
  • #7
morphism
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I've just had a thought, can I just say that as f(x)>0, then it cannot change the direction of the inequality, hence if a(x)<b(x) , then multiplying both sides by f(x), we find f(x)a(x)<f(x)b(x) and therefore

[tex]\int_{x_i}^{x_f} f(x)a(x) dx < \int_{x_i}^{x_f} f(x)b(x) dx [/tex]
That's perfect. (But you should be using [itex]\leq[/itex] instead of [itex]<[/itex].)

This is assuming that if a(x)<b(x), then

[tex]\int_{x_i}^{x_f} a(x) dx < \int_{x_i}^{x_f} b(x) dx [/tex]

But I don't understand why this second equation is true
This should be straightforward to prove if you just use the definition of the integral. To make it even easier, you can prove the following:

If c(x) is integrable on [x_i, x_f] and if c(x)>=0 for all x in [x_i, x_f], then

[tex]\int_{x_i}^{x_f} c(x) \geq 0.[/tex]

[Once you do this, you can put c(x)=b(x)-a(x) to get the result you want.]
 

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