Let f be a continuous function satisfying ∫−πt2 f(x)+x2dx=π+43t3 for all t, then fπ2/4 is equal to
π−π48
π2−π44
π2−π42
π−π416
Differentiating both sides, we have
ft2+t42t=4t2ft2+t4=2t⇒ ft2=2t−t4
so fπ24=fπ22=2π2−π24
=π−π416.