functions...........

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

f(1)=1

indeed f(x)=x