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

Solution :
We have
$$\eqalign{ & {R_1} = \int_{ - 1}^2 {x\,f\left( x \right)} dx = \int_{ - 1}^2 {\left( {1 - x} \right)f\left( {1 - x} \right)dx} \cr & \left[ {{\text{Using }}\int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)dx} } } \right] \cr & \Rightarrow {R_1} = \int_{ - 1}^2 {\left( {1 - x} \right)f\left( x \right)dx} \cr & \left[ {{\text{As}}\,{\text{ }}f\left( x \right) = f\left( {1 - x} \right){\text{ on }}\left[ { - 1,\,2} \right]} \right] \cr & \therefore {R_1} + {R_2} = \int_{ - 1}^2 {x\,f\left( x \right)dx + } \int_{ - 1}^2 {\left( {1 - x} \right)f\left( x \right)dx} \cr & \Rightarrow 2{R_1} = \int_{ - 1}^2 {f\left( x \right)dx = {R_2}} \cr} $$