
Integral inequality.
Hi,
I hope this is valid:
Why does
$\displaystyle \int_a^b s_n \le x_1 \le \int_a^b t_n$ and $\displaystyle \int_a^b s_n \le x_2 \le \int_a^b t_n$
multiply by 1 and add the result to the second give
$\displaystyle 0 \le x_2x_1 \le \int_a^b t_n  \int_a^b s_n$
and not
$\displaystyle \int_a^b s_n  \int_a^b t_n \le x_2x_1 \le \int_a^b t_n  \int_a^b s_n$
Thanks
Regards
Craig.

Hello,
Without further information, it's not possible to know that $\displaystyle x_2x_1\geq 0$

Hi,
I thought as much. This is from Apostol. I am sure that I have missed something. for example, $\displaystyle n \ge 1$, 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.