Understanding of swapping bounds of integral Please help me understand swapping the bounds of an integral better.
I learned that $$\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$$
Now when I try to visualize this, take $\sin(x)$ for example, $\int_{\pi}^{2\pi} \sin(x) dx$ and $- \int_{2\pi}^{\pi} \sin(x) dx$ both give answer $-2$, it somehow makes sense.
But when I try to visualize this, if I look at this part $\int_{2\pi}^{\pi} \sin(x) dx$ (without the minus sign), it gives me an answer to be 2, but visually when I go from  $2\pi$ to $\pi$, the area of $\sin(x)$ is still under x-axis. How do I interpret this?
 A: Here is a somewhat intuitive explanation.   
There is this concept of signed/orientated area. In the Euclidian place, if you are traversing the contour (is that the right word in English?) of the figure in counter-clockwise order, it's considered positive, I think. Otherwise it's considered negative.  
This is most simply illustrated when calculating triangle area.
Triangle Area
Check this part: 

"The (signed) area of a planar triangle specified by its vertices..."

So if you visit the vertices in a different order (w.r.t. clockwise or counterclockwise), you will get a different sign for the triangle area.
So in your case, you're traversing the contour in a different (counter-clockwise) order when looking from $2\pi$ to $\pi$.   
A: I guess it makes the most sense to use the fundamental theorem of calculus.
In this case
$$\int_{\pi}^{2\pi}\sin x\ dx=-\cos x|_{\pi}^{2\pi}=-\cos 2\pi+\cos \pi=-1-1=-2$$
and
$$\int_{2\pi}^{\pi}\sin x\ dx=-\cos x|_{2\pi}^{\pi}=-\cos{\pi}+\cos 2\pi=1+1=2\text{.}$$
As expected, they have the same value but opposite sign. I'm not sure there is a way to visualize this in a meaningful way. The way I have seen it, integrals are technically defined only with limits from $a$ to $b$ when $a\le b$. At some point, it becomes useful to define an integral from $b$ to $a$ as the negative of the integral from $a$ to $b$.
A: This can be seen best from the definition of an integral. 
For example we can see that $$-\int_{2\pi}^{\pi} sin(x)dx = \int_{2\pi}^{\pi} -sin(x)dx= \lim_{n\rightarrow \infty}\sum_{i=1}^{n}sin(x_i)(-\Delta x)$$
So the $\Delta x$ term is negative, i.e. you are moving from right to left, and thus if you didn't take the negative sign into account, your answer would be wrong.
A: Let me try without formulae. 
Imagine that integration is the sliding door opening the inner volume of a furniture. From left to right ($a$ to $b$), you open it. From right to left, you close it ($b$ to $a$). If you perform one operation, then the other, they cancel themselves. Hence, they should be of opposite sign.
It is the same when you take a step to the right, then to the left. No matter you have the "illusion" that you have performed twice the work, you are back to point $0$.
