Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,} $$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then
$${R_1} = \int\limits_{ - 1}^2 {xf(x)dx = \int\limits_{ - 1}^2 {(2 - 1 - x)f(2 - 1 - x)dx} } $$
$$ = \int\limits_{ - 1}^2 {(1 - x)f(1 - x)dx = \int\limits_{ - 1}^2 {(1 - x)f(x)dx} } $$
Hence, $$2{R_1} = \int\limits_{ - 1}^2 {f(x)dx = {R_2}} $$.
We can write the integral
$$\int\limits_0^b {{{(1 - x)}^2}dx - \int\limits_0^1 {{{(1 - x)}^2}dx = {1 \over 4}} } $$
$$ \Rightarrow \left. {{{{{(x - 1)}^3}} \over 3}} \right|_0^b - \left. {{{{{(x - 1)}^3}} \over 3}} \right|_b^1 = {1 \over 4}$$
$$ \Rightarrow {{{{(b - 1)}^3}} \over 3} + {1 \over 3} - \left( {0 - {{{{(b - 1)}^3}} \over 3}} \right) = {1 \over 4}$$
$$ \Rightarrow {{2{{(b - 1)}^3}} \over 3} = - {1 \over {12}} \Rightarrow {(b - 1)^3} = - {1 \over 8} \Rightarrow b = {1 \over 2}$$
$${x^2} = t \Rightarrow 2x\,dx = dt$$
$$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin t} \over {\sin t + \sin (\ln 6 - t)}}dt} $$ and $$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin (\ln 6 - t)} \over {\sin (\ln 6 - t) + \sin t}}dt} $$
$$2I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {1dt \Rightarrow I = {1 \over 4}\ln {3 \over 2}} $$.
