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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?

Mark Viola
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user321070
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  • So, this is true, no matter in which direction you integrate, the area of $\sin x$ from $\pi$ to $2\pi$ is under the $x$-axis. But if you start at $2\pi$, you're heading left-ward to get to $\pi$ as opposed to how we usually integrate right-ward. This is why there's a sign change embedded in the integration. – anonymouse Mar 08 '16 at 19:04
  • By definition of notation when $b<a$ $$\int_a^b f(x),dx=-\int_b^a f(x),dx$$when the latter integral exists. – Mark Viola Mar 08 '16 at 19:08
  • The question was asked in a different form in http://math.stackexchange.com/questions/1316529/why-does-an-integral-change-signs-when-flipping-the-boundaries – Laurent Duval Mar 08 '16 at 19:17

4 Answers4

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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 $\pi$.

peter.petrov
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2

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 $b$.

pancini
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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 $0$.

Laurent Duval
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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.

M47145
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