Mathematics Challenge Marathon

[3.14159265359]

Pretty much the fact that the function being undefined at 0, in fact asymptotically approaching 0 is heavily related to this problem. The integral is said to be divergent because the area under the graph is actually approaching \(\infty\) as \(x\to 0\), both in the positive and in the negative direction. The interval \(-2\leq x \leq 3\) clearly contains the value \(x=0\), so when we try to compute the area under the curve we end up crossing this border, which leads to big problems.

That's why the question was a bait. A student who only thinks in terms of antiderivatives and not in terms of what the integral actually represents would obtain the wrong answer.

However, it may be worth noting that not all functions that have asymptotes will have divergent integrals. For example, \( \int_{-2}^3 \frac{1}{x^{2/3}}\,dx \) converges (it equals to some finite real number). Wolfram won't give the exact correct answer for this since it combines complex analysis into the mixture and gives the principal value of \( (-2)^{1/3} \), which is complex, but plugging into this integral calculator will give a value that's clearly not undefined.

since at x=0 the graph is undefined, does that mean the area doesn't exist?

[RuiAce]

since at x=0 the graph is undefined, does that mean the area doesn't exist?

The thing is, interestingly enough the fact that \(x=0\) is undefined by itself does not imply that the area does not exist. The area certainly does exist, and can be computed by manually evaluating the integral.

This is more or less because \(x=0\) is a single point. If the function was undefined on a whole interval instead, then the notion of area falls apart entirely. But if the function is undefined at a single point, the area may or may not exist. This is why mathematicians study convergence - it helps us investigate under what conditions should that area exist and pose no problems.