Chapter 6.2, Problem 4E

To determine

To express the given limit as a definite integral.

Answer to Problem 4E

  ${\displaystyle \underset{2}{\overset{3}{\int }}\frac{1}{1-x}dx}$

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:

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$

Calculation:

  $\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{1}{1-c_k}\Delta x},\left[2,3\right]\\ \Rightarrow f\left(c_k\right)=\frac{1}{1-c_k}\\ \Rightarrow f\left(x\right)=\frac{1}{1-x}\end{array}$

Therefore definite integral is,

  ${\displaystyle \underset{2}{\overset{3}{\int }}f\left(x\right)dx}={\displaystyle \underset{2}{\overset{3}{\int }}\frac{1}{1-x}dx}$