My Double Integrals!

$\displaystyle\int_0^\frac{1}{2}\left (\displaystyle\int_0^{1-2y} xy \mathrm dx\right ) \mathrm dy\\\displaystyle\int_0^\frac{1}{2}\left [y\frac{x^2}{2}\right ]_0^{1-2y} \mathrm dy\\ \displaystyle\int_0^\frac{1}{2} y\frac{(1-2y)^2}{2}\mathrm dy\\ \displaystyle\int_0^\frac{1}{2} \left (\frac{y}{2}-2y^2+2y^3\right )\mathrm dy\\\left [\frac{y^2}{4}-\frac{2y^3}{3}+\frac{y^4}{2}\right ]_0^\frac{1}{2}\\\frac{1}{16}-\frac{1}{12}+\frac{1}{32}=\frac{1}{96}=\frac{a}{b}\\a=1,b=96\Rightarrow a+b=97$