Chapter 5.9, Problem 1E

a.

To determine

To find $L_2,R_2\text{and}{M}_2$ .

Answer to Problem 1E

The value of $L_2,R_2\text{and}{M}_2$ is 6, 12 and 9.6.

Explanation of Solution

Given information: $I={\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}$

Concept Used:

1. To find the area of $n$ rectangles( $L_n$) by using the formula as:

  $\begin{array}{l}{L}_n=\Delta x\left[f(x_0)+f(x_1)+\mathrm{.........}+f(x_{n-1})+\right]\\ \text{where,}\Delta x=\frac{b-a}{n}\end{array}$

2. To find the area of $n$ rectangles( $R_n$) by using the formula as:

  $\begin{array}{l}{R}_n=\Delta x\left[f(x_1)+f(x_2)+\mathrm{.........}+f(x_n)\right]\\ \text{where,}\Delta x=\frac{b-a}{n}\end{array}$

3. The midpoint rule is defined as:

  $\begin{array}{l}{M}_n=\Delta x[f(\overline{x_1})+f(\overline{x_2})+\mathrm{.....}+f(\overline{x_n})]\\ \text{where,}\Delta x=\frac{b-a}{n}\\ \text{and}\overline{x_i}=\frac{1}{2}\left(x_{i-1}+x_i\right)\end{array}$

Here, $n=2,a=0,b=4,\Delta x=2,f(0)=0.5,f(1)=1.5,f(2)=2.5,f(3)=3.3,f(4)=3.5$

  $L_2=2\left[f(x_0)+f(x_1)\right]=2(0.5+2.5)=6$

  $R_2=2\left[f(x_1)+f(x_2)\right]=2(2.5+3.5)=12$

  $M_2=2\left[f(\overline{x_1})+f(\overline{x_2})\right]=2[f(1)+f(3)]=2(1.5+3.3)=9.6$

Hence, the value of $L_2,R_2\text{and}{M}_2$ is 6, 12 and 9.6

b.

To determine

To test whether the values of $L_2,R_2\text{and}{M}_2$ are underestimates or overestimates of $I$ .

Answer to Problem 1E

$L_2$ underestimates and $R_2,M_2$ overestimates $I$ .

Explanation of Solution

Given information: $I={\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}$

The value of $L_2,R_2\text{and}{M}_2$ is 6, 12 and 9.6

Since, $L_2<<I$ , the value underestimates of $I$ .

and $R_2,M_2>>I$ , the value overestimates the value of $I$ .

c.

To determine

To estimate the value of $I$ by finding $T_2$ .

Answer to Problem 1E

The value of $T_2=9$ .

Explanation of Solution

Given information: $I={\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}$

Concept Used:

Trapezoidal rule is given by the formula as:

  $\begin{array}{l}{T}_n=\frac{\Delta x}{2}[(f(x_0)+f(x_n)+2(f(x_1)+\mathrm{........}+f(x_{n-1})]\\ \text{where,}\Delta x=\frac{b-a}{n}\end{array}$

$T_n=\frac{2}{2}[(f(0)+f(4)+2f(2)]=0.5+3.5+2(2.5)=9$

Hence, the value of $T_2=9$ .

d.

To determine

To find the value of $n$ , $L_n,R_n,T_n,I\text{and}{M}_n$ are in increasing order.

Answer to Problem 1E

For all $n$ , $L_n<T_n<I<M_n<R_n$

Explanation of Solution

Given information: $I={\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}$

If the value of $n$ is increased, the curve becomes flat but the order of the values remains unchanged.

For all $n$ , $L_n<T_n<I<M_n<R_n$