Question:
If $f(a+b+1-x)=f(x)$, for all $x$, where $a$ and $b$ are fixed positive real numbers,
then $\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) d x$ is equal to:
Correct Option: , 3
Solution:
$I=\frac{1}{(a+b)} \int_{a}^{b} x[f(x)+f(x+1)] d x$
$x \rightarrow a+b-x$
$I=\frac{1}{(a+b)} \int_{a}^{b}(a+b-x)[f(a+b-x)+f(a+b+1-x)] d x$
$I=\frac{1}{(a+b)} \int_{a}^{b}(a+b-x)[f(x+1)+f(x)] d x$ ....(ii)
$[\because$ put $x \rightarrow x+1$ in $f(a+b+1-x)=f(x)]$
Add (i) and(ii)
$2 I=\int_{a}^{b}[f(x+1)+f(x)] d x$
$2 I=\int_{a}^{b} f(x+1) d x+\int_{a}^{b} f(x) d x$
$=\int_{a}^{b} f(a+b+1-x) d x+\int_{a}^{b} f(x) d x$
$2 I=2^{b} \int_{a}^{b} f(x) d x$
$\therefore \int_{a-1}^{b-1} f(x+1) d x$ $[\because$ Put $x \rightarrow x+1]$