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.

Your question is,
$$\dfrac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$$
Now, in second term i.e. $f(x+1)$, put $x+1$ as $u$ and use $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ and $f(a+b+1 -x) = f(x)$.
You will get $$\dfrac{1}{a+b} \int_{a}^{b} (a+b) \left[ f(x) \right] dx$$. Now substitute $x\rightarrow x+1$, you'll get option 3.
