0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

$\dfrac1{x^n}\to0$ 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 ?).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.