If f(a+b+1-x)=f(x),


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:

  1. (1) $\int_{a+1}^{b+1} f(x) d x$

  2. (2) $\int_{a-1}^{b-1} f(x) d x$

  3. (3) $\int_{a-1}^{b-1} f(x+1) d x$

  4. (4) $\int_{a+1}^{b+1} f(x+1) d x$

Correct Option: , 3


$I=\frac{1}{(a+b)} \int_{a}^{b} x[f(x)+f(x+1)] d x$

$x \rightarrow a+b-x$

$I=\frac{1}{(a+b)} \int_{a}^{b}(a+b-x)[f(a+b-x)+f(a+b+1-x)] d x$

$I=\frac{1}{(a+b)} \int_{a}^{b}(a+b-x)[f(x+1)+f(x)] d x$ ....(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 $x \rightarrow x+1]$


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