Question

Consider the improper integral \( I = \int_{2025}^{2030} \frac{1}{(x - 2025)^k} \, dx \), where \( k > 0 \). Which of the following is true for \( I \)?

• A. \( I \) is convergent if \( k<1 \) and is divergent if \( k \geq 1 \)
• B. \( I \) is convergent if \( k \leq 1 \) and is divergent if \( k>1 \)
• C. \( I \) is convergent if \( k \geq 1 \) and is divergent if \( k<1 \)
• D. \( I \) is convergent if \( k>1 \) and is divergent if \( k \leq 1 \)

Hint:

For improper integrals of the form \( \int_{0}^{a} x^{-k} \, dx \), the integral converges if \( k<1 \) and diverges if \( k \geq 1 \).

Answer 1

The Correct Option is A

We are given an improper integral: \[ I = \int_{2025}^{2030} \frac{1}{(x - 2025)^k} \, dx \] Let \( u = x - 2025 ⇒ x = u + 2025 \), then when \( x = 2025 \), \( u = 0 \) and when \( x = 2030 \), \( u = 5 \). So the integral becomes: \[ I = \int_{0}^{5} \frac{1}{u^k} \, du \] This is an improper integral because the integrand becomes unbounded at \( u = 0 \). To analyze convergence at the lower limit, consider the behavior of the integral: \[ \int_{0}^{5} \frac{1}{u^k} \, du \] This integral converges if and only if \( k<1 \), and diverges if \( k \geq 1 \). Hence, the original integral \( I \) converges for \( k<1 \) and diverges for \( k \geq 1 \).