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I realize there's a bunch of similar questions, but for derivative. However, this is a little bit different.

I understand pretty well why derivative of sin(x) is cos(x) and of cos(x) is -sin(x). That makes sense, since derivative expresses the "angle of normal", I can see that from graph easily.

However, when I try to - analogically, using a graph - find integral of sin(x), I can't seem to make sense of it.

Here's my graphs:

I know that integral is "area under graph". If you look at graph of sin(x), it starts with zero area under the graph and gradually more is added, with growing speed, then it slows down and eventually stops at PI, where the derivative reaches zero. Then as the graph goes under the X axis, the area is subtracted again and eventually it reaches zero at 2PI.

That doesn't quite add up with -cos(x), does it? It would make sense if it was as it's on my 3rd picture.

I probably understand it wrong, so please give me some hint how to understand it... the point is, I never remember the formulas, and I'm reasonably successful with finding the derivatives, but never get the integral right.

When you talk about integrals, you could be talking about one of many distinct, but related topics. You could mean Riemann integrals, indefinite integrals, or many other types of integral out there.

You're probably talking about the indefinite integral when you say that the integral of $sin$ is $-cos$. The thing about the indefinite integral is that it is not really a single function. That's why you often display it with the little ${}+C$ at the end. An indefinite integral is more like a set of functions, which are all equal under vertical translations.

So, $int f$ could be considered the set of functions which have a derivative $f$. There are many functions which have the derivative $sin$, and they are all of the form $-cos+C$, where $C$ is some real number. This is why it is said the integral of $sin$ is $-cos$.

But why does the integral then not represent the area? Well, it does. What you were saying was that $-cos(x)$ should be the area under $sin(x)$ up to $x$. The question is, up to $x$ from where? There is no lower bound! If you pick the lower bound $0$, then

$$-cos(x)-(-cos(0))=-cos(x)+1$$

You get that $+1$ you were wondering about, and it does represent the area under $sin(x)$ from $0$ to $x$.

You are right as far as area goes but remember that the integral is only defined up to a constant!

Numerically if you were directly comparing areas then the function you would need would be $1-cos{x}$. However recall that this is indefinite integration.