If f(a+b+1-x)=f(x) for all x, where a and b a | vedclass

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(D) Let $I = \frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) dx \quad \dots(1)$Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we get:$

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$\int_{a-1}^{b-1} f(x) dx$

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(D) Let $I = \frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) dx \quad \dots(1)$Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we get:$I = \frac{1}{a+b} \int_{a}^{b} (a+b-x)(f(a+b-x)+f(a+b-x+1)) dx$Given $f(a+b+1-x) = f(x)$,we have $f(a+b-x) = f(x+1)$.Thus,$I = \frac{1}{a+b} \int_{a}^{b} (a+b-x)(f(x+1)+f(x)) dx \quad \dots(2)$Adding $(1)$ and $(2)$:$2I = \frac{1}{a+b} \int_{a}^{b} (a+b)(f(x)+f(x+1)) dx = \int_{a}^{b} (f(x)+f(x+1)) dx$Since $f(a+b+1-x) = f(x)$,let $t = a+b+1-x$,then $dt = -dx$. When $x=a, t=b+1$; when $x=b, t=a+1$.$\int_{a}^{b} f(x) dx = \int_{a+1}^{b+1} f(a+b+1-t) dt = \int_{a+1}^{b+1} f(t) dt$.Thus,$2I = \int_{a}^{b} f(x) dx + \int_{a}^{b} f(x+1) dx = \int_{a+1}^{b+1} f(x+1) dx + \int_{a}^{b} f(x+1) dx$. This simplifies to $I = \int_{a}^{b} f(x+1) dx = \int_{a-1}^{b-1} f(x) dx$.

If $f(a+b+1-x)=f(x)$ for all $x,$ where $a$ and $b$ are fixed positive real numbers,then $\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) dx$ is equal to:

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