Let f be a polynomial function such that f(x2+1)=x4+5x2+2, for all x∈ℝ. Then ∫03f(x) dx is equal to: [2026]
(2)
∵ f(x2+1)=x4+5x2+2
Put x2+1=t
⇒f(t)=(t-1)2+5(t-1)+2
⇒f(t)=t2+3t-2
Now, ∫03f(t) dt=∫03(t2+3t-2) dt
=[t33+3t22-2t]03
=[273+272-6]
=332