Chapter 6.5, Problem 16E

(a)

To determine

To calculate: The value of the integral using Simpson’s rule with n=4 approximation

Answer to Problem 16E

The value of integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx}=0.69325$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_1^2\frac{1}{x}dx}$

Formula used:

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{.....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Calculation:

To approximate ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{.....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Where [ a , b ] is partitioned into an even number n of subintervals of equal length $h=\frac{b-a}{n}$

Now, apply the Simpson’s rule with

  $\begin{array}{l}n=4\\ f\left(x\right)=\frac{1}{x}\\ [a,b]=[1,2]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{2-1}{4}\\ h=\frac{1}{4}\end{array}$

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+y_4\right)}\\ =\frac{\frac{1}{4}}{3}\left(1+4\left(\frac{4}{5}\right)+2\left(\frac{2}{3}\right)+4\left(\frac{4}{7}\right)+\frac{1}{2}\right)\\ =\frac{1}{12}\left(1+\frac{16}{5}+\frac{4}{3}+\frac{16}{7}+\frac{1}{2}\right)\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx}=0.69325\end{array}$

Conclusion:

The value of integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx}=0.69325$

(b)

To determine

To calculate: The exact value of the integral is ${\displaystyle {\int }_1^2\frac{1}{x}dx}$

Answer to Problem 16E

The value of the integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx=0.69315}$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_1^2\frac{1}{x}dx}$

Formula used:

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

Calculation:

If f is continuous at every point of [a, b], and if F is any anti-derivative of f on [a, b], then

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

This part of the fundamental theorem is also called the integral evaluation theorem.

Now, use the above formula to find the value of integral.

  $f\left(x\right)=\frac{1}{x}$

If F is any anti-derivative of f then

  $F\left(x\right)=\ln x$

So,

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx=\left(\ln 2-\ln 1\right)}\\ =0.693-0\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx=0.69315}\end{array}$

Conclusion:

The value of the integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx=0.69315}$