If f(a+b+1-x)=f(x), for all x, where a and b are fixed positive real numbers, then 1a+b_a^…

Question

If $f(a+b+1-x)=f(x)$, for all $x$, where $a$ and $b$ are fixed positive real numbers, then $\dfrac1{a+b}\displaystyle\int_a^bx(f(x)+f(x+1))\,dx$ is equal to:

Accepted Answer

$$I=\frac1{a+b}\int_a^bx[f(x)+f(x+1)]dx\ \ldots(i)$$

Using $x\to a+b-x$:
$$I=\frac1{a+b}\int_a^b(a+b-x)[f(a+b-x)+f(a+b+1-x)]dx=\frac1{a+b}\int_a^b(a+b-x)[f(x+1)+f(x)]dx\ \ldots(ii)$$

(using $f(a+b+1-x)=f(x)$, then $x\to x+1$ inside gives $f(a+b-x)=f(x+1)$)

Adding (i) and (ii):
$$2I=\int_a^b[f(x+1)+f(x)]dx=\int_a^bf(x+1)dx+\int_a^bf(x)dx=\int_a^bf(a+b+1-x)dx+\int_a^bf(x)dx=2\int_a^bf(x)dx$$

$$\therefore I=\int_a^bf(x)\,dx=\int_{a-1}^{b-1}f(x+1)\,dx\quad(\text{shifting }x\to x+1)$$

Solutions · 1