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?
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.