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I was exploring integrals, and I found something interesting that I couldn't prove.

$$\int_0^\pi \sin(x)dx = 2$$

If you have ever seen the $\sin(x)$ or $\cos(x)$ function, you would notice it is wavy and makes "bumps" alternating positive and negative. Well, it turns out that the area of each of those "bumps" is equal to 2. Here is a visualization:

The area I am interested in

The reason this stood out to me was because I didn't expect it to be 2. I expected some crazy random number, but I got 2!

I'm assuming there some sort of rule in trigonometry that can explain this. Can anyone help?

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    You can *compute* it as $-\cos \pi + \cos 0 = 2$ because $(-\cos)$ is an antiderivative of $\sin$. – 2017-01-20
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    Since sin(x) can be seen as the height of a circle at some point, then the integral from pi to 0 can be thought of as 'how much the height changes'. From 0, the sin function goes up to a maximum of 1, then goes back down to 0 as the angle goes to pi. The 'distance' it traveled is 2. – 2017-01-20
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    @Malcolm: No, that would be $\int_0^\pi |\frac{d}{dx}\sin(x)|\,dx$ -- which also happens to be $2$, but for quite different reasons. – 2017-01-20
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    If you just kind of stare at it, it's not unreasonable. Draw the vertical line at 2. On what remains, you can imagine cutting off a little tail piece and putting it in the top right corner of the box formed by the horizontal line at 1 and the vertical line at 2,and then just turning the rest over and just putting it in. Of course, it wouldn't be perfect, and this is far from a 'mathematical' answer as to why it 'should' be 2. On the other hand, our spatial intuition says that 2 is a perfectly good number for this value to be. The rest is just a coincidence of integration. – 2017-01-20

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It's not clear to me whether you did the integral yourself or just plugged it into a computer, so I'll give the straightforward explanation: an integral is done by taking the antiderivative of the function and taking the difference of the values of the endpoints. It turns out that the derivative of $\cos(x)$ is $-\sin(x)$, so the antiderivative of $\sin(x)$ is $-\cos(x)$. Thus $\int_0^{\pi}\sin(x)dx = -\cos(\pi) - -\cos(0) = 1 - -1 = 2$.

Now, that's not very satisfying - I assume you're interested in some sort of geometric argument for why that area "should" be $2$. To my knowledge, there is no such argument - there's no useful symmetry to take advantage of, for example. It's like asking why the square root of $1024$ is $32$; it just is.

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we have $$\int_{0}^{\pi}\sin(x)dx=-\cos(x)|_{x=0}^{x=\pi}=-\cos(\pi)-(-\cos(0))=1+1=2$$