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