## duality function!

$\displaystyle f(x,y) = kx(x-y), 0<x<1, -x<y<x$

this is how i approached it;

$\displaystyle \int_{-x}^{x}\int_{0}^{1}kx(x-y)dxdy = 1$

$\displaystyle \int_{-x}^{x}\left[k(\frac{x^3}{3} - yx)\right]_{0}^{1}dy = 1$

$\displaystyle \left[k(\frac{y}{3} - \frac{y^2}{2}\right]_{-x}^{x} = 1$

$\displaystyle \left[k(\frac{x}{3} - \frac{x^2}{2}) - (\frac{-x}{3} - \frac{(-x)^2}{2})\right] = 1$

reducing it further, i have

$\displaystyle k(\frac{2x}{3}) = 1$

$\displaystyle k = \frac{3x}{2}$

it is ok to stop here and write the function

$\displaystyle f(x,y) = \left[\frac{3x^2}{2}(x-y), 0<x<1, -x<y<x$$\displaystyle 0 , elsewhere \right]$

or something else to do, thanks