Chapter 5.9, Problem 7E

a.

To determine

To find : the approximate value of integral with the specified value of n .

Answer to Problem 7E

the approximate value of integral by trapezoidal rule is $2.41195$ .

Explanation of Solution

Given information :

The integral is ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx,n=8$ .

Formula used : trapezoidal rule:

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=T_n=\frac{\Delta x}{2}[f\left(x_0\right)+2f(x_1)+2f\left(x_2\right)+\mathrm{...}+2f\left(x_{n-1}\right)+f\left(x_n\right)]$

Where $\Delta x=\frac{\left(b-a\right)}{n}\text{ and }{x}_i=a+i\Delta x$ .

Calculation : by using trapezoidal rule,

  ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=T_8$

Since $a=0,b=2$ and $n=8$ .

So,

  $\begin{array}{l}\Delta x=\frac{\left(b-a\right)}{n}\\ \Delta x=\frac{\left(2-0\right)}{8}\text{ [substitute]}\\ \Delta x=\frac{1}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_0=0+0\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_0=0\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_1=0+1\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_1=\frac{1}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_2=0+2\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_2=\frac{2}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_3=0+3\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_3=\frac{3}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_4=0+4\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_4=\frac{4}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_5=0+5\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_5=\frac{5}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_6=0+6\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_6=\frac{6}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_7=0+7\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_7=\frac{7}{4}\text{ [simplify]}\end{array}$

  $\begin{array}{l}{x}_i=a+i\Delta x\\ x_8=0+8\left(\frac{1}{4}\right)\text{ [substitute]}\\ x_8=\frac{8}{4}\text{ [simplify]}\end{array}$

Here $f\left(x\right)=\sqrt[4]{1+x^2}$ ,

So

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_0\right)=\sqrt[4]{1+{\left(0\right)}^2}\text{ [substitute }{x}_0=0]\\ f\left(x_0\right)=1\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_1\right)=\sqrt[4]{1+{\left(\frac{1}{4}\right)}^2}\text{ [substitute }{x}_1=\frac{1}{4}]\\ f\left(x_1\right)=1.015\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_2\right)=\sqrt[4]{1+{\left(\frac{2}{4}\right)}^2}\text{ [substitute }{x}_2=\frac{2}{4}]\\ f\left(x_2\right)=1.057\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_3\right)=\sqrt[4]{1+{\left(\frac{3}{4}\right)}^2}\text{ [substitute }{x}_3=\frac{3}{4}]\\ f\left(x_3\right)=1.117\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_4\right)=\sqrt[4]{1+{\left(\frac{4}{4}\right)}^2}\text{ [substitute }{x}_4=\frac{4}{4}]\\ f\left(x_4\right)=1.189\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_5\right)=\sqrt[4]{1+{\left(\frac{5}{4}\right)}^2}\text{ [substitute }{x}_5=\frac{5}{4}]\\ f\left(x_5\right)=1.264\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_6\right)=\sqrt[4]{1+{\left(\frac{6}{4}\right)}^2}\text{ [substitute }{x}_6=\frac{6}{4}]\\ f\left(x_6\right)=1.3423\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_7\right)=\sqrt[4]{1+{\left(\frac{7}{4}\right)}^2}\text{ [substitute }{x}_5=\frac{7}{4}]\\ f\left(x_7\right)=1.417\text{ [simplify]}\end{array}$

  $\begin{array}{l}f\left(x\right)=\sqrt[4]{1+x^2}\\ f\left(x_8\right)=\sqrt[4]{1+{\left(\frac{8}{4}\right)}^2}\text{ [substitute }{x}_8=\frac{8}{4}]\\ f\left(x_8\right)=1.493\text{ [simplify]}\end{array}$

Now the value is

  $\begin{array}{l}{\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=\frac{\Delta x}{2}[f\left(x_0\right)+2f(x_1)+2f\left(x_2\right)+2f\left(x_3\right)+2f\left(x_4\right)+2f\left(x_5\right)+2f\left(x_6\right)+2f\left(x_7\right)+f\left(x_8\right)]\\ {\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=\frac{1}{4·2}[1+2·1.015+2·1.057+2·1.117+2·1.189+2·1.264+2·1.3423+2·1.417+1.493]\text{ [substitute]}\\ {\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=\frac{1}{8}[19.2956]\text{ [add]}\\ {\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=2.41195\text{ [divide]}\end{array}$

So the approximate value of integral by trapezoidal rule is $2.41195$ .

b.

To determine

To find : the approximate value of integral with the specified value of n .

Answer to Problem 7E

the approximate value of integral by midpoint rule is $2.4475$ .

Explanation of Solution

Given information :

The integral is ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx,n=8$ .

Formula used:

Midpoint rule:

  $\begin{array}{l}{\displaystyle {\int }_a^{b}f\left(x\right)dx=M_n=\Delta x[f\left({\overline{x}}_1\right)}+f\left({\overline{x}}_2\right)+\mathrm{.....}+f\left({\overline{x}}_n\right)]\\ \Delta x=\frac{b-a}{n}\\ {\overline{x}}_i=\frac{1}{2}\left(x_{i-1}+x_i\right)=\text{midpoint of }[x_{i-1},x_i]\end{array}$

Calculation: by using midpoint rule,

  ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=M_8$

Since $a=0,b=2$ and $n=8$ .

So,

  $\begin{array}{l}\Delta x=\frac{\left(b-a\right)}{n}\\ \Delta x=\frac{\left(2-0\right)}{8}\text{ [substitute]}\\ \Delta x=\frac{1}{4}\text{ [simplify]}\end{array}$

The midpoint of the eight subintervals are $\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{9}{8},\frac{11}{8},\frac{13}{8},\frac{15}{8}$ .

So by the midpoint rule:

  ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}dx=\Delta x[f\left({\overline{x}}_1\right)}+f\left({\overline{x}}_2\right)+f\left({\overline{x}}_3\right)+f\left({\overline{x}}_4\right)+f\left({\overline{x}}_5\right)+f\left({\overline{x}}_6\right)+f\left({\overline{x}}_7\right)+f\left({\overline{x}}_8\right)]$

  ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}dx=\frac{1}{4}[f\left(\frac{1}{8}\right)}+f\left(\frac{3}{8}\right)+f\left(\frac{5}{8}\right)+f\left(\frac{7}{8}\right)+f\left(\frac{9}{8}\right)+f\left(\frac{11}{8}\right)+f\left(\frac{13}{8}\right)+f\left(\frac{15}{8}\right)]\text{ [substitute]}$

  ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}dx=\frac{1}{4}[}\sqrt[4]{1+{\left(\frac{1}{8}\right)}^2}+\sqrt[4]{1+{\left(\frac{3}{8}\right)}^2}+\sqrt[4]{1+{\left(\frac{5}{8}\right)}^2}+\sqrt[4]{1+{\left(\frac{7}{8}\right)}^2}+\sqrt[4]{1+{\left(\frac{9}{8}\right)}^2}+\sqrt[4]{1+{\left(\frac{11}{8}\right)}^2}+\sqrt[4]{1+{\left(\frac{13}{8}\right)}^2}+\sqrt[4]{1+{\left(\frac{15}{8}\right)}^2}]$

  ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}dx=\frac{1}{4}[}1+1.017+1.029+1.085+1.148+1.22+1.30+1.33+1.45]\text{ [simplify]}$

  $\begin{array}{l}{\displaystyle {\int }_0^2\sqrt[4]{1+x^2}dx=\frac{1}{4}[}10.579]\text{ [add]}\\ {\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=2.4475\text{ [divide]}\end{array}$

So, the approximate value of integral by midpoint rule is $2.4475$ .

c.

To determine

To find : the approximate value of integral with the specified value of n .

Answer to Problem 7E

the approximate value by simpson’s rule is $2.4101$ .

Explanation of Solution

Given information :

The integral is ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx,n=8$ .

Formula used:

Simpson’s rule:

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=S_n=\frac{\Delta x}{3}[f\left(x_0\right)+4f(x_1)+2f\left(x_2\right)+\mathrm{...}+4f\left(x_{n-1}\right)+f\left(x_n\right)]$

Where n is even and $\Delta x=\frac{\left(b-a\right)}{n}$ .

Calculation: by using simpson’s rule,

  ${\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=S_8$

Since $a=0,b=2$ and $n=8$ .

So,

  $\begin{array}{l}\Delta x=\frac{\left(b-a\right)}{n}\\ \Delta x=\frac{\left(2-0\right)}{8}\text{ [substitute]}\\ \Delta x=\frac{1}{4}\text{ [simplify]}\end{array}$

By simpson’s rule,

  $\begin{array}{l}{\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=\frac{\Delta x}{3}[f\left(x_0\right)+4f(x_1)+2f\left(x_2\right)+4f\left(x_3\right)+2f\left(x_4\right)+4f\left(x_5\right)+2f\left(x_6\right)+4f\left(x_7\right)+f\left(x_8\right)]\\ {\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=\frac{1}{12}[1+4·1.015+2·1.057+4·1.117+2·1.189+4·1.264+2·1.3423+4·1.417+1.493]\text{ [substitute]}\\ {\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=\frac{1}{12}[28.9216]\text{ [add]}\\ {\displaystyle {\int }_0^2\sqrt[4]{1+x^2}}dx=2.4101\text{ [divide]}\end{array}$

Thus, the approximate value by simpson’s rule is $2.4101$ .