Chapter 5.2, Problem 4E

To determine

To express: the given limit as a definite integral.

Answer to Problem 4E

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{1}{1-c_k}\Delta x}={\displaystyle \underset{2}{\overset{3}{\int }}\frac{1}{1-x}dx}$ .

Explanation of Solution

Given:

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{1}{1-c_k}\Delta x},\left[2,3\right]$ .

Concept used:

Let $f$ be a continuous function on $\left[a,b\right]$ , and let $\left[a,b\right]$ be partitioned into $n$ disjoint intervals of equal length $\Delta x=\frac{b-a}{n}$ . Then the definite integral of $f$ over $\left[a,b\right]$ is given by $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}$ , where each $c_k$ is chosen arbitrarily in the $k\text{th}$ subinterval.

Calculation:

Consider the given limit,

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{1}{1-c_k}\Delta x},\left[2,3\right]$

Comparing with $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}$ , it follows that

  $f\left(c_k\right)=\frac{1}{1-c_k}$

So, $f\left(x\right)=\frac{1}{1-x}$ .

Hence,

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{1}{1-c_k}\Delta x}={\displaystyle \underset{2}{\overset{3}{\int }}\frac{1}{1-x}dx}$