Question

Let $f:[-1,2]$ $\to [0,\infty ]$ be a continuous function such that
$f(x)=f(1-x)forallx\in [-1,2].$ If $R_1=\underset{-1}{\overset{2}{\int }}xf(x)dx$ 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)

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