# Is this inequality true?

1. Aug 8, 2008

### jenga42

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

2. Aug 8, 2008

### snipez90

Your notation is a bit awkward. Are f(x)a(x) and f(x)b(x) the integrands?

3. Aug 8, 2008

### CompuChip

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
$$| \int_a^b f(x) p(x) dx | < | f(x) p(x) (b - a) | < \cdots$$

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

4. Aug 8, 2008

### morphism

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

$$\int_a^b h(x) dx \leq \int_a^b k(x) dx.$$

Edit: Oh. Maybe that's what you were saying here:

5. Aug 8, 2008

### jenga42

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

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

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

for a specific f(x).

What does "the integral is monotone" mean?

Thanks!

6. Aug 8, 2008

### jenga42

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

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

????

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

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

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. Aug 8, 2008

### morphism

That's perfect. (But you should be using $\leq$ instead of $<$.)

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

$$\int_{x_i}^{x_f} c(x) \geq 0.$$

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