If $$fa+b+1$$−$$x=fx$$, for all x, where a and b are fixed positive real numbers, then $$1a+b$$∫$$abxfx+fx+1dx$$ is equal to

Correct option is 1∫a+1b+1fxdx

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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

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detailed solution

Correct option is A

fx+1=fa+b−xI=1a+b∫abxfx+fx+1dx........1I=1a+b∫aba+b−xfx+1+fxdx......2From 1 and 2 2I=∫abfx+fx+1dx2I=∫abfa+b−xdx+∫abfx+1dx2I=2∫abfx+1dx⇒I=∫abfx+1dx=∫a+1b+1fxdx

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If fa+b+1−x=fx, for all x, where a and b are fixed positive real numbers, then 1a+b∫abxfx+fx+1dx is equal to

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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