Official solution
I = \[{1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx\] ...(1)
x \[ \to \] a + b - x
I = \[{1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {a + b - x} \right) + f\left( {a + b - x + 1} \right)} \right)dx} \]
I = \[{1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} \] .....(2)
[As f(x) = f(a + b + 1 - x)
\[ \Rightarrow \] f(x + 1) = f(a + b - x)]
Adding (1) and (2) we get
2I = \[{{a + b} \over {a + b}}\int\limits_a^b {\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} \]
\[ \Rightarrow \] I = \[{1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {x + 1} \right)dx} \]
\[ \Rightarrow \] I = \[{1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {a + b - x + 1} \right)dx} \]
\[ \Rightarrow \] I = \[{1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( x \right)dx} \]
\[ \Rightarrow \] I = \[\int\limits_a^b {f\left( x \right)dx} \]
Let x = t + 1
\[ \therefore \] I = \[\int\limits_{a - 1}^{b - 1} {f\left( {t + 1} \right)dt} \]
= \[\int\limits_{a - 1}^{b - 1} {f\left( {x + 1} \right)dx} \]