Chapter 5.2, Problem 27E

To determine

To express the integral as a limit of Riemann sum.

Answer to Problem 27E

The value of Riemann sum is $\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[\frac{\left(2+\frac{4}{n}i\right)}{1+{\left(2+\frac{4}{n}i\right)}^5}\right]}\frac{4}{n}$ .

Explanation of Solution

Given information:

The equation is

  ${\int }_2^6\frac{x}{1+x^5}dx$

The given expression can be obtained as

  ${\int }_2^6\frac{x}{1+x^5}dx$

The width of the subintervals is

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{6-2}{n}\left[\text{since, }b=6,a=2\right]\\ \Delta x=\frac{4}{n}\end{array}$

And

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_i=2+\frac{4}{n}i\left[\text{since, }a=2,\Delta x=\frac{4}{n}\right]\end{array}$

Therefore,

  ${\int }_2^6\frac{x}{1+x^5}dx=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$

Put the values of $\Delta x=\frac{4}{n},x_i=2+\frac{4}{n}i$ on the above equation

  $\begin{array}{l}{\int }_2^6\frac{x}{1+x^5}dx=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}\\ \text{ =}\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[\frac{\left(2+\frac{4}{n}i\right)}{1+{\left(2+\frac{4}{n}i\right)}^5}\right]}\frac{4}{n}\end{array}$

Hence,

The value of Riemann sum is $\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[\frac{\left(2+\frac{4}{n}i\right)}{1+{\left(2+\frac{4}{n}i\right)}^5}\right]}\frac{4}{n}$ .