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

$2R_1 = R_2$

We have,

$R_1=\int\limits_{-1}^{2}xf(x) dx$

Using: $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x) dx$, we obtain

$R_1=\int\limits_{-1}^{2}(1-x)(1-x)dx$

$⇒R_1=\int\limits_{-1}^{2}(1-x)f(x)dx$   $∵f(1-x)=f(x)(given)$

$⇒R_1=\int\limits_{-1}^{2}f(x)dx-R_1⇒2R_1=\int\limits_{-1}^{2}f(x)dx$

It is given that $R_2 = \int\limits_{-1}^{2}f (x) dx$

$∴2R_1 = R_2$