Change this integral $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$. It is given that $f(a+b+1 -x) = f(x)$ where $a$ and $b$ are positive real numbers then $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$ is equal to

*

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


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


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


*$\int_{a+1}^{b+1} f(x) dx$
I tried using integration by parts along with this property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ and the one given in the question but I'm unable to get anything. It's that extra $x$ and the fraction $\frac{1}{a+b}$ which is making everything quite uneasy.

$$I=\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx=I_1+I_2$$
$$I_1=\frac{1}{a+b}\int_{a}^{b}xf(x) dx$$
$$I_2=\frac{1}{a+b} \int_{a}^{b} x f(x+1) dx = \frac{1}{a+b} \int_{a+1}^{b+1} (u-1)f(u) du$$
N0w apply the property:
$$\int_{p}^{q} g(x) dx=\int_{p}^{q} g(p+q-x) dx$$
Then $$I_2=\frac{1}{a+b} \int_{a+1}^{b+1} (a+1+b+1-u-1)f(a+b+2-u) du$$
Use $f(a+b+2-u)=f(a+b+1-(u-1))= f(u-1)$. then
$$I_2=\frac{1}{a+b} \int_{a+1}^{b+1} (a+b+1-u) f(u-1) du.$$
Let $u-1=w$, then $$I_2=\frac{1}{a+b} \int_{a}^{b} (a+b-w)f(w) dw$$
Finally $$I=I_1+I_2=\int_{a}^{b} f(w) dw =\int_{a-1}^{b-1} f(x+1)dx$$
