Apparent paradox in the use of certain integration formulas 
background
here is a formula which we all know in integral:
$$
\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx
$$
and this, too:
$$
\int_{a}^{b}f(x)dx=\int_{a}^{\frac{b+a}{2}}f(x)+f(a+b-x)dx
$$
and we also know, in  graphic view, these formulas could work because the $f(x)\space$ and the  $f(a+b-x)\space$ are symmetric to each other in $[a,b]$ and the axis of their symmetry is $x=\frac{a+b}{2}$.
so that we could  'fold them up'  to a smaller range and calculate it both.

question body statement
now, what if we 'fold up' the range again?
$$let \space b_{1}=\frac{a+b}{2},\space f_{1}(x)=f(x)+f(a+b-x)$$
and we get these below:
$$\int_{a}^{b_{1}}f_{1}(x)dx=\int_{a}^{b_{1}}f_{1}(2b_{1}-x)dx
$$
$$
\int_{a}^{b_{1}}f_{1}(x)dx=\int_{a}^{\frac{b_{1}+a}{2}}f_{1}(x)+f_{1}(2b_{1}-x)dx
$$
they also could work, and $\int_{a}^{b_{1}}f_{1}(x)dx=\int_{a}^{b}f(x)dx$,right?
suppose above equations could make sense,
then we keep on doing this procedure, what will happen?
first,we define some variables
$$b_{n}=\frac{a+b_{n-1}}{2},b_{0}=b,
$$
$$
f_{n}(x)=f_{n-1}(x)+f_{n-1}(2b_{n-1}-x),
$$
$$
I_{n}=\int_{a}^{b_{n}}f_{n}(x)dx
$$
as a result, we get a series:
$$\{I_{n}\}$$
as noted before, we also have a eternal equation:
$$\{I_{n}\}\equiv\int_{a}^{b}f(x)dx$$

here comes the paradox!
if we let $n$ trend to $\infty$,we would get:
$$
\lim_{n\rightarrow\infty}I_{n}=\int_{a}^{b}f(x)dx
$$
obviously ,right?
but, we also get:
$$
\lim_{n\rightarrow\infty}b_{n}=a
$$
that means:
$$
\lim_{n\rightarrow\infty}I_{n}=\int_{a}^{b_{n}}f_{n}(x)dx=\int_{a}^{a}f_{n}(x)dx=0
$$
why? how could this happen? 
in graphic view, it is obvious, after infinite folding , the  area of the zone between the $f(x)$ and y=0 will trend to zero,
but in algebra view, this shall not happen.
how to explain the paradox?
 A: Actually, in the graphical view it is not obvious that "after infinite folding" the area under the curve is zero.
Let's take a very simple example: a constant function, $f(x) = 1.$
Then 
$$f_1(x) = f(x) + f(a+b−x) = 1 + 1 = 2.$$
So in going from $\int_a^b f(x)\,dx$ to $\int_a^{b_1} f_1(x)\,dx$
we have cut the horizontal distance in half
($b_1 - a = \frac12(b - a)$)
but we have doubled the height of the graph.
Hence the area remains the same.
Fold again and you will have $\frac14$ as much horizontal distance,
but $4$ times as much height.
For non-constant functions you will usually find that $f_1 \neq 2f,$
that is, the function is not exactly doubled everywhere, 
but the average height of $f_1$ over $[a,b_1]$ will be twice the average height of $f$ over $[a,b].$
After $n$ folds you have a function $f_n$ whose average height over
$[a,b_n]$ is $2^n$ times the average height of $f$ over $[a,b].$
In other words, no area is ever lost under the graph, it just gets piled up higher and closer to the $y$-axis with every fold.
In order to sketch a graph of the integral at the limit, $b_\infty=a,$
you would have to somehow plot a function $f_\infty$ whose value on $[a,a]$ is exactly $2^\infty$ times the average value of $f$ over $[a,b].$
There is no such function in real analysis, but if you assume (incorrectly) that there is such a function that you can integrate on $[a,a],$
you will conclude that the integral is zero.
So one  mistake is assuming that there is any meaning at all to the integral
$\int_a^{b_\infty} f_\infty(x)\,dx,$ either arithmetically or graphically.
There is no graphical interpretation of what the integral would look like
"after infinite folding."
But there is actually another mistake: you assume that
you can evaluate a limit by jumping to the limiting case.
Consider this false "proof":
\begin{align}
1 &= \frac12 + \frac12 \\
  &= \frac14 + \frac14 + \frac14 + \frac14 \\
  &= \frac18 + \frac18 + \frac18 + \frac18
       + \frac18 + \frac18 + \frac18 + \frac18 \\
  & \qquad\vdots \\
  &= \frac{1}{2^n} + \frac{1}{2^n} + \cdots + \frac{1}{2^n}
        \quad \text{($2^n$ terms)} \\
  & \qquad\vdots \\
  &= 0 + 0 + 0 + \cdots \\
  &= 0.
\end{align}
The fallacy here is in the jump from a finite sum of non-zero terms
to the sum $0+0+0+\cdots$.
There's simply no justification for that step.
