A question from a multivariable calculus exam: I have tried lots of methods like integrating the RHS by parts. Any help would be appreciated.
Find $w(y)$ such that the identity $$ \intop_{-1}^{1} \intop_{0}^{x}f(y)dydx=\intop_{0}^{1}w(y)[f(y)-f(-y)]dy $$ holds for every integrable function $f:[-1,1]\mapsto\mathbb{R}.$
Thanks for the suggestions, does this look right:
Answer:
LHS =
$$ \intop_{-1}^{1}\intop_{0}^{x}f(y)dydx=\intop_{-1}^{0}\intop_{y}^{-1}f(y)dxdy+\intop_{0}^{1}\intop_{y}^{1}f(y)dydx $$
$$ =\intop_{-1}^{0}f(y)(-1-y)dy+\intop_{0}^{1}f(y)(1-y)dy $$
integral w.r.t. $x$ is just constant
$$ =\intop_{0}^{1}f(y)(1+y)dy+\intop_{0}^{1}f(y)(1-y)dy $$
$$ =\intop_{0}^{1}f(-z)(1-z)(-dz)+\intop_{0}^{1}f(y)(1-y)dy\text{ making the substitution $z=-y$} $$
$$ =-\intop_{0}^{1}f(-y)(1-y)(dy)+\intop_{0}^{1}f(y)(1-y)dy\ $$
since z is dummy variable
$$ =\intop_{0}^{1}(1-y)[f(y)\mathbf{-}f(-y)]dy $$
So,
$$ w(y)=(1-y) $$