If p(x) be a function defined on R such that p'(x)=p'(1-x), ∀x∈[0,1],p(0)=1and p(1)=41. Then, ∫01p(x)dx
41
21
42
Explanation for the correct option.
Step 1: Find px
Given that, p'(x)=p'(1-x),x∈[0,1],p(0)=1,p(1)=41.Integrating both sides in p'(x)=p'(1-x) we have:
p(x)=-p(1-x)+cAt x=0,p(0)=-p(1)+c1=-41+cc=42
Step 2: Integrate ∫01p(x)dxI=∫01p(x)dx...(1)I=∫01p(1-x)dx..(2)Adding (1) and (2)2I=∫01p(x)+p(1-x)dx=42∫01dx[p(x)+p(1-x)=42]2I=42I=21
Hence, option B is correct.