Consider an improper integral
∫
1
∞
1
𝑥
2
𝑎
−
1
𝑑
𝑥
.

Which condition should
𝑎
satisfy to make this improper integral convergent?

Select the best answer (which includes as many values as possible).

Question

Consider an improper integral 
∫
1
∞
1
𝑥
2
𝑎
−
1
𝑑
𝑥
.

Which condition should  
𝑎
  satisfy to make this improper integral convergent?

Select the best answer (which includes as many values as possible).

BlackTom AI · Accepted Answer

We are given the improper integral ∫ from 1 to ∞ of 1/x^(2a−1) dx, and we need to determine for which values of a this integral converges. Option 1: a > 1 — This matches the standard convergence criterion for integrals of the form ∫_1^∞ x^(-p) dx, which converges if and only if p > 1. Here p = 2a − 1. Requiring 2a − 1 > 1 yields 2a > 2, so a > 1. This option aligns with the correct condition for convergence. Option 2: a > 1 (written as |a| > 1) — The absolute value condition would include negative a with large magnitude (e.g., a = −2 gives exponent 2a − 1 = −5, making the integrand x^5, which certainly does not converge on [1, ∞)). Moreover, the convergence criterion depends on 2a − 1 > 1, which does not translate to |a| > 1. Hence this broad absolute-value condition is incorrect. Option 3: a > 0.5 — If a = 0.75, then the exponent is 2(0.75) − 1 = 0.5, and the integral ∫_1^∞ x^(-0.5) dx diverges. Therefore a > 0.5 is not sufficient for convergence, making this option incorrect. Option 4: a < 1 — This directly contradicts the required a > 1 and would fail for values that would otherwise yield convergence. Thus it is not correct. Option 5: 0 < a < 1 — This interval gives 2a − 1 between −1 and 1, so the exponent p = 2a − 1 is ≤ 1, which does not satisfy p > 1. Therefore the integral diverges in this range, making this option incorrect. Option 6: a ≥ 1 — This includes a = 1, which yields exponent 2(1) − 1 = 1, giving ∫_1^∞ 1/x dx, a divergent integral. Thus a ≥ 1 is too inclusive and not correct. Putting the pieces together, only the condition a > 1 ensures that the exponent 2a − 1 exceeds 1, which is necessary for convergence of the integral on [1, ∞).

MAT137Y1 LEC 20249: Calculus with Proofs (all lecture sections) Pre-Class Quiz 54 (12.1-12.6)

Consider an improper integral ∫ 1 ∞ 1 𝑥 2 𝑎 − 1 𝑑 𝑥 . Which condition should 𝑎 satisfy to make this improper integral convergent? Select the best answer (which includes as many values as possible).

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Similar Questions

Let 𝑓 , 𝑔 be positive, continuous functions with domain 𝑅 . Let 𝐿 = lim 𝑥 → ∞ 𝑓 ( 𝑥 ) 𝑔 ( 𝑥 ) . Assume ∫ 1 ∞ 𝑓 ( 𝑥 ) 𝑑 𝑥 is convergent. Which of the following statements must be true? Select all the correct answers.

Let 𝑓 be a positive, continuous function with domain 𝑅 . Assume ∫ 1 ∞ 𝑓 ( 𝑥 ) 𝑑 𝑥 is divergent. Which of the following improper integrals must also be divergent? Select all the correct answers.

Let 𝑓 be a positive, continuous function with domain 𝑅 . Assume ∫ 1 ∞ 𝑓 ( 𝑥 ) 𝑑 𝑥 is convergent. Which of the following improper integrals must also be convergent? Select all the correct answers.

Which of the following integrals are IMPROPER? Select all the correct answers.

Which of the following statements is correct? Note: only one of them is correct.

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

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