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

Let us denote

$$I=\frac1{a+b}\int_{a}^{b}x\bigl(f(x)+f(x+1)\bigr)\,dx.$$

To evaluate the numerator, we split it into two separate integrals:

$$\int_{a}^{b}x\bigl(f(x)+f(x+1)\bigr)\,dx =\underbrace{\int_{a}^{b}x\,f(x)\,dx}_{J} +\underbrace{\int_{a}^{b}x\,f(x+1)\,dx}_{K}.$$

We first examine the second part, $$K$$. The functional relation given in the question is

$$f(a+b+1-x)=f(x)\qquad\text{for every }x.$$

Replacing $$x$$ by $$x+1$$ in this relation gives

$$f\!\bigl(a+b+1-(x+1)\bigr)=f(x+1) \;\;\Longrightarrow\;\; f(a+b-x)=f(x+1).$$

Hence

$$K=\int_{a}^{b}x\,f(x+1)\,dx =\int_{a}^{b}x\,f(a+b-x)\,dx.$$

Now make the change of variable $$t=a+b-x\;(\Rightarrow\,x=a+b-t,\;dx=-dt)$$. When $$x=a$$, $$t=b$$; when $$x=b$$, $$t=a$$. Therefore

$$\begin{aligned} K&=\int_{x=a}^{x=b}x\,f(a+b-x)\,dx \\[4pt] &=\int_{t=b}^{t=a}(a+b-t)\,f(t)\,(-dt) \\[4pt] &=\int_{a}^{b}(a+b-t)\,f(t)\,dt. \end{aligned}$$

The first part is simply

$$J=\int_{a}^{b}t\,f(t)\,dt,$$

where we have renamed the dummy variable $$x$$ to $$t$$ for later convenience.

Adding $$J$$ and $$K$$ gives

$$\begin{aligned} J+K&=\int_{a}^{b}t\,f(t)\,dt+\int_{a}^{b}(a+b-t)\,f(t)\,dt \\[4pt] &=\int_{a}^{b}\bigl[t+(a+b-t)\bigr]f(t)\,dt \\[4pt] &=\int_{a}^{b}(a+b)\,f(t)\,dt \\[4pt] &=(a+b)\int_{a}^{b}f(t)\,dt. \end{aligned}$$

Returning to $$I$$ we have

$$I=\frac1{a+b}\bigl(J+K\bigr) =\frac1{a+b}\,(a+b)\int_{a}^{b}f(t)\,dt =\int_{a}^{b}f(t)\,dt.$$ Replacing the dummy variable $$t$$ by $$x$$ once again,

$$I=\int_{a}^{b}f(x)\,dx.$$

It remains to relate this result to the options given. Consider Option A:

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

Make the substitution $$u=x+1\;(\!\Rightarrow x=u-1,\,dx=du)$$.

When $$x=a-1$$, $$u=a$$; when $$x=b-1$$, $$u=b$$. Thus

$$\int_{a-1}^{\,b-1}f(x+1)\,dx =\int_{u=a}^{u=b}f(u)\,du =\int_{a}^{b}f(u)\,du =\int_{a}^{b}f(x)\,dx.$$

This is exactly the value we obtained for $$I$$. Hence the required expression equals Option A.

Hence, the correct answer is Option A.