I'm studying improper integrals with Paul's Online Notes as a reference. Sorry if I'm quoting it here, but the website has the following problem:
Determine if the following integral is convergent or divergent. If it is convergent find its value. $$\int_{-2}^{3} \frac{1}{x^3} \, dx$$
And the solution provided in the website is:
This integrand is not continuous at $x=0$ and so we'll need to split the integral up at that point. $$\int_{-2}^{3} \frac{1}{x^3} \, dx=\int_{-2}^0\frac{1}{x^3}\,dx+\int_{0}^3\frac{1}{x^3}\,dx$$ Now we need to look at each of these integrals and see if they are convergent. $$\int_{-2}^0\frac{1}{x^3}\,dx=\lim_{t \to 0^-}\int_{-2}^t\frac{1}{x^3}\,dx$$ $$=\lim_{t \to 0^-}(-\frac{1}{2t^2}+\frac{1}{8})$$ $$=-\infty$$ At this point we're done. One of the integrals is divergent that means the integral that we were asked to look at is divergent. We don't even need to bother with the second integral.
Question is is the solution correct? If I use my intuition, the integral $$\int_{-2}^{3} \frac{1}{x^3} \, dx$$ should be equal to $$\int_{2}^{3} \frac{1}{x^3} \, dx$$ because $$\int_{-2}^{2} \frac{1}{x^3} \, dx=0$$ If it's true, then the integral should be convergent and its value should be $\frac{5}{72}$.
I checked an online integral calculator https://www.integral-calculator.com/ and it seemed to confirm my answer. So which solution and reasoning is correct and why?