Given equation is ∫0t2(f(x)+x2)dx=34t3,∀t>0
According to Newton Leibnitz theorem we havedxd(∫u(x)v(x)f(t)dt)=f(v(x))×v′(x)−f(u(x))×u′(x)
Apply Newtons Leibnitz theorem in the given equation.
⇒(f(t2)+t4)2t−0=4t2
⇒f(t2)+t4=2t
⇒f(x2)=−x4+2x
⇒f(x)=−x2+2x
⇒f(4π2)=−42π4+2×2π
=16−π4+π
=π(1−16π3)
Hence this is the correct option.