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

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- Oct 19th 2010, 12:04 AMSambitintegration problem
$\displaystyle \iint\limits_R \cos [\frac{\pi x^2}{2}]dxdy$=?, where R is a region bounded by y=0, x=1, y=x.

- Oct 19th 2010, 01:46 AMPandevil1990
- Oct 19th 2010, 03:18 AMmr fantastic
- Oct 19th 2010, 04:56 AMSambit
@ 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 AMSambit
ok...so here is what i have done when

**"[ ]" means box-function**....

$\displaystyle 0 < x < 1 ==> 0 < \frac{\pi x^2}{2} < \frac{\pi}{2} = 1.57 $(approx.)

==>$\displaystyle [\frac{\pi x^2}{2}]=0, if \frac{\pi x^2}{2} < 1, $ ie, $\displaystyle x < \sqrt\frac{2}{\pi}$

so the integration becomes $\displaystyle \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.