Answer:
Option C
Explanation:
Here R1=∫2−1xf(x)dx
Using, ∫baf(x)dx=∫baf(a+b−x)dx
R1=∫2−1(1−x)f(1−x)dx,
[∵f(x)=f(1−x)]
∴ R1=∫2−1(1−x)f(x)dx.........(ii)
Given , R2 is area bounded by
f(x),x-1 and x=2
∴ R2==∫2−1f(x)dx ...(iii)
Adding Eqs.(i) and (ii) , we get
2R==∫2−1f(x)dx......(iv)
∴ From Eqs.(iii) and (iv) , we get
2R1=R2