# Definition of an integral

1. Jun 23, 2011

I have seen in some textbooks that the value of integrals are treated differently.

In one textbook it said to evaluate
$$\int_{-2}^{2} x^3 dx$$
I would say immediately that the answer is zero. because the function is odd around origo.

While the textbook claimed you had to split ut the integral, and integrate term by term.

I would have agreed with the textbook if it had said , find the area of x^3 from -1 to 1. But just giving the integral, isnt this wrong?

2. Jun 23, 2011

### micromass

You are certainly correct in saying that the integral is zero because the function is odd. I don't quite see why the textbook should say that. I guess the textbook wants you to find the area, and then you need to split up the integral. However, if they just give the integral and say nothing about the area, then you are correct in saying that it is 0.

So, judging from what you give me, I would say that you are correct.

3. Jun 23, 2011

### Stephen Tashi

I wonder if the textbook was treating the topic of whether
$\int_{-\infty}^{\infty} f(x) dx$ can be evaluated as $\lim_{a \rightarrow \infty} \int_{-a}^{a} f(x) dx$.

4. Jun 24, 2011

### HallsofIvy

I hope not. That statement is not true!
$$\int_{-\infty}^\infty f(x)dx$$
must be evaluated as
$$\lim_{\epsilon\to -\infty}\int_\epsilon^a f(x)dx+ \lim_{\delta\to\infty} \int_a^\delta f(xt)dx$$
where the two limits are evaluated independently. The formula you give is the "Cauchy principle value" which is the same as the integral if the integral exist but may exist even when the integral itself does not.

5. Jun 24, 2011

### Stephen Tashi

Why do you hope not? If the textbook was treating that topic then Nebuchadnezza wasn't hallucinating (at least not to a large extent).