i think Y should be f(x).
if it be correct,then take ∫y dy=a &∫Y2 dy= b ; with limits 0 to 1
then f(x)=(1+b)x+ ax2 ;
now solve for a & b,
f(1)=1+a+b,
the calculation is very hectic
f(x)=x+∫01(x2y+y2x)dy
find f(1)(x and y are independent variables)
since,
∫(x2 y+y2 x) dy =x/3+x2/3 (limit 0→1)
f(x)=x+x/3+x^2/3
=8x+3x2/6
f(1)=11/6
ABSOLUTELY WRONG.................................................
i think Y should be f(x).
if it be correct,then take ∫y dy=a &∫Y2 dy= b ; with limits 0 to 1
then f(x)=(1+b)x+ ax2 ;
now solve for a & b,
f(1)=1+a+b,
the calculation is very hectic