4 multiple choice 1 point the table above shows several riemann sum approximations to ∫₀¹(1/x)dx using right…

4 multiple choice 1 point the table above shows several riemann sum approximations to ∫₀¹(1/x)dx using right - hand endpoints of n subintervals of equal length of the interval 0,1. which of the following statements best describes the limit of the riemann sums as n approaches infinity? (a) the limit of the riemann sums is infinity? (b) the limit of the riemann sums is a finite number less than 10. (c) the limit of the riemann sums does not exist because (1/xₙ)(1/n) does not approach 0. (d) the limit of the riemann sums does not exist because it is a sum of infinitely many positive numbers. (e) the limit of the riemann sums does not exist because ∫₀¹(1/x)dx does not exist. a b c d e previous next

4 multiple choice 1 point the table above shows several riemann sum approximations to ∫₀¹(1/x)dx using right - hand endpoints of n subintervals of equal length of the interval 0,1. which of the following statements best describes the limit of the riemann sums as n approaches infinity? (a) the limit of the riemann sums is infinity? (b) the limit of the riemann sums is a finite number less than 10. (c) the limit of the riemann sums does not exist because (1/xₙ)(1/n) does not approach 0. (d) the limit of the riemann sums does not exist because it is a sum of infinitely many positive numbers. (e) the limit of the riemann sums does not exist because ∫₀¹(1/x)dx does not exist. a b c d e previous next

Answer

Answer:

A. The limit of the Riemann sums is infinity.

Explanation:

Step1: Recall improper - integral definition

The integral $\int_{0}^{1}\frac{1}{x}dx$ is an improper integral of Type 1 (discontinuity at $x = 0$). The Riemann sum $\sum_{k = 1}^{n}\left(\frac{1}{x_{k}}\right)\left(\frac{1}{n}\right)$ with right - hand endpoints on $[0,1]$ is an approximation of $\int_{0}^{1}\frac{1}{x}dx$.

Step2: Evaluate the improper integral

We know that $\int_{0}^{1}\frac{1}{x}dx=\lim_{a\rightarrow0^{+}}\int_{a}^{1}\frac{1}{x}dx$. Using the antiderivative of $\frac{1}{x}$ which is $\ln|x|$, we have $\lim_{a\rightarrow0^{+}}[\ln(1)-\ln(a)]=\lim_{a\rightarrow0^{+}}(-\ln(a))=\infty$.

Step3: Analyze the limit of Riemann sums

As $n$ approaches infinity, the Riemann sums $\sum_{k = 1}^{n}\left(\frac{1}{x_{k}}\right)\left(\frac{1}{n}\right)$ approximate the value of the improper integral $\int_{0}^{1}\frac{1}{x}dx$. Since the improper integral $\int_{0}^{1}\frac{1}{x}dx$ diverges to infinity, the limit of the Riemann sums as $n$ approaches infinity is infinity.