Chapter 6.5, Problem 17E

(a)

To determine

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

Answer to Problem 17E

The value of integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx}=5.2522$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^4\sqrt{x}dx}$

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

  $\begin{array}{l}n=4\\ f\left(x\right)=\sqrt{x}\\ [a,b]=[0,4]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{4-0}{4}\\ h=1\end{array}$

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{x}dx=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+y_4\right)}\\ =\frac{1}{3}\left(0+4\left(1\right)+2\left(\sqrt{2}\right)++4\left(\sqrt{3}\right)+2\right)\\ =\frac{1}{3}\left(0+4+2.828+6.928+2\right)\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx}=5.2522\end{array}$

Conclusion:

The value of integral is ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{x}dx}=5.2522$

(b)

To determine

To calculate: The exact value of the integral ${\displaystyle {\int }_0^4\sqrt{x}dx}$

Answer to Problem 17E

The value of the integral is ${\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{x}dx=5.333}$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^4\sqrt{x}dx}$

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.

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{x}dx=\left(\frac{2}{3}\times 4^{\frac{3}{2}}-\frac{2}{3}\times 0^{\frac{3}{2}}\right)}\\ =5.333-0\\ {\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{x}dx=5.333}\end{array}$

Conclusion:

The value of the integral is ${\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{x}dx=5.333}$