Discussion Forum

Let $f:[-1,2] \rightarrow [0, \infty]$ be a continuous function such that f(x) = f(1 - x) for all, $x\epsilon [-1,2]$. Let $R_{1}=\int_{-1}^{2} x f(x) dx$ and $R_{2}$  be the area of the region bounded by y = f(x),x = - 1, x = 2and the X-axis. Then

Answer:

Explanation:

Here  $R_{1}= \int_{-1}^{2} xf(x) dx$

 Using, $\int_{a}^{b} f(x) dx=\int_{a}^{b}  f(a+b-x) dx$

 $R_{1}=\int_{-1}^{2} (1-x)f(1-x) dx, $ 

                                           $[ \because   f(x)=f(1-x)]$

 $\therefore$      $R_{1}=\int_{-1}^{2} (1-x)f(x) dx$.........(ii)

 Given , $R_{2}$ is area bounded by

 f(x),x-1 and x=2

 $\therefore$   $R_{2}==\int_{-1}^{2} f(x) dx$ ...(iii)

 Adding Eqs.(i) and (ii) , we get

  $2R= =\int_{-1}^{2} f(x) dx$......(iv)

$\therefore$   From Eqs.(iii) and (iv) , we get

  $2R_{1}=R_{2}$