Question

Let $f : [-1,2]$ $ \rightarrow [0, \infty] $ be a continuous function such that
$f(x) = f ( 1 - x)\, for \,all\, x \in [-1,2].$ If $ R_1 = \int \limits_{-1}^2 x f(x) dx $ and $ R_2 $ are the area of the region bounded by $y = f(x), x = - 1 , x = 2$ and the $x-axis$. Then,

• A. $(a)R_1 = 2R_2 $
• B. $(b)R_1 = 3R_2 $
• C. $(c)2R_1 = R_2 $
• D. $(d)3R_1 = R_2 $

Answer 1

Answer: (C)

$ R_1 = \int \limits_{-1}^2 x f(x) dx \, \, \, \, \, .......(i) $
Using $\, \, \, \, \int \limits_a^b f(x) dx = \int \limits_a^b f(a + b - x) dx $
$\, \, \, \, \, \, \, R_1 = \int \limits_{-1}^2 (1 - x) f(1 - x) dx $
$ \therefore \, \, \, \, \, \, R_1 = \int \limits_{-1}^2 (1 - x) f(x) dx $
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, [f(x) = f(1 - x), given] $
Given, $ R_2 $ is area bounded by f(x), x = - 1 and x = 2.
$ \therefore \, \, \, \, \, \, \, \, \, \, R_1 = \int \limits_{-1}^2 f(x) dx \, \, \, ...(iii) $
On adding Eqs. (i) and (ii), we get
$ \, \, \, \, \, \, \, \, \, \, 2R_1 = \int \limits_{-1}^2 f(x) dx\, \, \, \, \, \, \, ....(iv) $
From Eqs. (iii) and (iv), we get
$ \, \, \, \, \, \, \, \, \, \, 2R_1 = R_2 $