# integration problem

• Oct 19th 2010, 12:04 AM
Sambit
integration problem
$\iint\limits_R \cos [\frac{\pi x^2}{2}]dxdy$=?, where R is a region bounded by y=0, x=1, y=x.
• Oct 19th 2010, 01:46 AM
Pandevil1990
• Oct 19th 2010, 03:18 AM
mr fantastic
Quote:

Originally Posted by Sambit
$\iint\limits_R \cos [\frac{\pi x^2}{2}]dxdy$=?, where R is a region bounded by y=0, x=1, y=x.

The first step is to reverse the order of integration (Fubini's Theorem).
• Oct 19th 2010, 04:56 AM
Sambit
@ Pandevil1990,

what you have done is absolutely right...thanks for that. but here the "[ ]" means box-function....so what will be the answer in that case?
• Oct 21st 2010, 06:19 AM
Sambit
ok...so here is what i have done when "[ ]" means box-function....

$0 < x < 1 ==> 0 < \frac{\pi x^2}{2} < \frac{\pi}{2} = 1.57$(approx.)
==> $[\frac{\pi x^2}{2}]=0, if \frac{\pi x^2}{2} < 1,$ ie, $x < \sqrt\frac{2}{\pi}$

so the integration becomes $\int_0^{\sqrt{\frac{2}{\pi}}}\int_0^x cos [\frac{\pi x^2}{2}]dy dx + \int_{\sqrt{\frac{2}{\pi}}}^1\int_0^x cos [\frac{\pi x^2}{2}]dy dx

= \int_0^{\sqrt{\frac{2}{\pi}}}\int_0^x cos 0 dy dx + \int_{\sqrt{\frac{2}{\pi}}}^1\int_0^x cos 1 dy dx
= \frac{1}{\pi} + cos 1(\frac{1}{2} - \frac{1}{\pi})$

this is all i could do. but this answer does not agree with any of the the given possible options.