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, ∞).

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). Single choice

Similar Questions

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