Questions about hyperbolas and integration

Question

I have a couple of questions regarding hyperbolas and their integrals. If it's too much, don't feel like you have to answer all 3 questions.

My first question:

The integral of a function like 1/x^2 is -1/x - I sufficiently understand the algebraic manipulations that gets us to this result, but geometrically (and in terms of intuitive understanding) I am at 0. How can the integral of such a function (which is always above 0) be a negative number?

Is it due to the area between -1 and 1, so that you go "beyond" infinity and you end up with a negative - sort of like the idea that 1+2+3+4+… = -1/12 (the true logic behind which I of course can't appreciate).

The clue that makes me think all this is that as you add more and more positive area (as x increases for the indefinite integral of 1/x^2), the positive additions reduce the "infinite" negativity (that exists, as theorized, due to the first bit of the function around x=0) , hence why -1/x approaches 0. To get to 0, you would have to take the indefinite integral at x=infinity, which can be geometrically "explained". All this mumbo jumbo is also somewhat consistent with the fact that the definite integral of such a function (1/x^n) is a positive number as F(b) - F(a) is a positive (thank god)!

My second question:

What is the definite integral of 1/x^2 between x=1 and x=infinity related to (pi^2)/6 (the basel problem.

Finally (and most importantly):

Does calculus eventually become second nature and perfectly intuitive like arithmetic?

Answer 1

This isn't meant as a complete answer, but is just too long to fit nicely in a comment.

1. For the first, you said it yourself. The 'area' under the graph is precisely the definite integral, so it being positive is certainly consistent. The primitive is negative where $x>0$, but that's what you'd expect from a decreasing function, isn't it?

2. For the second, you can find the integral using the primitive you indicated. It isn't precisely related to $\sum_{k\geq 1}\frac{1}{k^2}$, except that both imply by their convergence that the other converges as well.

3. (Yes. At least this stuff does. At least I felt like it did.)

Answer 2

$\dfrac1{x^n}\to0$ as $x\to\infty$, for positive n. Hence, if the primitive were positive, then, by subtracting something positive from $0$, you'd really get a negative number, and then things would really make no sense, since then you'd really have a sum of positive quantities yielding a negative result! But, by being negative, we have $-(-a)=+a>0$ for positive a, which obviously makes perfect sense! If, for instance, $\displaystyle\int_a^bf(x)dx$ were defined as $F(a)-F(b)$, instead of the other way around, then the primitive would have to be positive... only then we'd have $\displaystyle\int x^ndx=-\frac{x^{n+1}}{n+1}$. :-) So basically, the idea is that the order of subtraction has to fit with whether the function is either increasing or decreasing. (Is this clear or intuitive enough ?).