In Question 0, select all of the integrals that DIVERGE.

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We are given a multiple-answer question about which integrals diverge, along with the labeled options and the provided answer indicating that (a) and (c) are the ones that diverge.

Option (a): Consider the general criteria for divergence of an improper integral. An integral diverges if the limit defining the integral does not exist or if the integrand grows without bound in a way that makes the area under the curve infinite, such as a singularity inside the interval or improper behavior at infinity. Since the answer key marks (a) as diverging, we would expect either a non-integrable singularity within the interval or an improper limit that does not converge to a finite value when evaluated.

Option (b): In this scenario, the key indicates that (b) does not diverge, implying that the integral is convergent. A common way this happens is if the integrand is integrable over the interval, either because it is bounded and the interval is finite, or because it decays sufficiently fast at infinity or near a singularity to yield a finite value. If a student were checking (b), they would look for a finite limit or a standard convergence test that confirms convergence, such as a comparison to a known convergent p-integral, or a finite improper limit.

Option (c): Like (a), (c) is marked as diverging by the key. Therefore, we should expect similar diagnostic signals of divergence: either a non-removable singularity within the interval, or an improper behavior at an endpoint (infinite bound, or a limit that does not converge). The reasoning would involve identifying such a feature and showing that it prevents the integral from attaining a finite value.

To understand why (b) would not diverge, one can compare the integrand (for (b)) to a known convergent case or apply an appropriate test. For instance, if the integral is over a finite interval and the integrand is continuous there, the integral is finite. If the interval is infinite, a decay condition like |f(x)| ≤ C/x^p with p > 1 often ensures convergence. The key contrast is that (b) lacks the features that cause (a) and (c) to diverge, according to the provided answer.

In summary, the divergence status hinges on the presence of non-integrable singularities or improper limits for (a) and (c), and the absence of such features for (b). The given answer aligns with this interpretation by marking (a) and (c) as divergent and (b) as non-divergent, which would be consistent with standard divergence/convergence criteria for improper integrals.

MAT136 - Calculus 2 - Winter 2025 Week 6 Tutorial Submission

In Question 0, select all of the integrals that DIVERGE.

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