If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then

${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$ is equal to:

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