# Integral inequality.

• Jun 23rd 2009, 08:25 AM
craigmain
Integral inequality.
Hi,

I hope this is valid:
Why does

$\int_a^b s_n \le x_1 \le \int_a^b t_n$ and $\int_a^b s_n \le x_2 \le \int_a^b t_n$

multiply by -1 and add the result to the second give

$0 \le x_2-x_1 \le \int_a^b t_n - \int_a^b s_n$

and not

$\int_a^b s_n - \int_a^b t_n \le x_2-x_1 \le \int_a^b t_n - \int_a^b s_n$

Thanks
Regards
Craig.
• Jun 23rd 2009, 11:19 AM
Moo
Hello,

Without further information, it's not possible to know that $x_2-x_1\geq 0$
• Jun 23rd 2009, 08:57 PM
craigmain
Hi,

I thought as much. This is from Apostol. I am sure that I have missed something. for example, $n \ge 1$, and these integrals are the upper and lower step function bounds. It's part of the integral definition.

I suppose because the upper step function is always above the lower step function that the difference must be more than zero.

Maybe I need to read it more carefully.