# Chapter 07.08: Simpson’s 3/8 th Rule of Integration

## Learning Objectives

After successful completion of this lesson, you should be able to:

1)  derive the formula for Simpson’s 3/8 rule of integration,

2)  use Simpson’s 3/8 rule it to solve integrals,

3)  develop the formula for composite (also called multiple-segment) Simpson’s 3/8 rule of integration,

4)  use composite Simpson’s 3/8 rule of integration to solve integrals,

5)  compare true error formulas for composite Simpson’s 1/3 rule and composite Simpson’s 3/8 rule, and

6)  use a combination of Simpson’s 1/3 rule and Simpson’s 3/8 rule to approximate integrals.

## Introduction

The main objective of this chapter is to develop appropriate formulas for approximating the integral of the form

$I = \int_{a}^{b}{f(x)dx}\ \ \ (1)$

Most (if not all) of the developed formulas for integration are based on a simple concept of approximating a given function $$f(x)$$ by a simpler function (usually a polynomial function) $$f_{i}(x)$$, where $$i$$ represents the order of the polynomial function. In Chapter 07.03, Simpsons 1/3 rule for integration was derived by approximating the integrand $$f(x)$$ with a 2nd order (quadratic) polynomial function.$$f_{2}(x)$$

$f_{2}(x) = a_{0} + a_{1}x + a_{2}x^{2}\ \ \ (2)$ Figure 1 $$\widetilde{f}(x)$$ Cubic function.

In a similar fashion, Simpson 3/8 rule for integration can be derived by approximating the given function$$f(x)$$ with the 3rd order (cubic) polynomial $$f_{3}(x)$$

$\begin{split} \left. \ \begin{matrix} f_{3}(x) &= a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} \\ &= \left\{ 1,x,x^{2},x^{3} \right\} \times \begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \\ \end{bmatrix} \\ \end{matrix} \right\}\ \ \ (3) \end{split}$

which can also be symbolically represented in Figure 1.

Method 1

The unknown coefficients $$a_{0},\ a_{1},\ a_{2}\ \text{and} \ a_{3}$$ in Equation (3) can be obtained by substituting 4 known coordinate data points $$\{ x_{0},f\left( x_{0} \right)\},\{ x_{1},f\left( x_{1} \right)\},\{ x_{2},f\left( x_{2} \right)\}\ \text{and}\ \{ x_{3},f\left( x_{3} \right)\}$$ into Equation (3) as follows.

$\left. \ \begin{matrix} f(x_{0}) = a_{0} + a_{1}x_{0} + a_{2}x_{0}^{2} + a_{3}x_{0}^{2} \\ f(x_{1}) = a_{0} + a_{1}x_{1} + a_{2}x_{1}^{2} + a_{3}x_{1}^{2} \\ f(x_{2}) = a_{0} + a_{1}x_{2} + a_{2}x_{2}^{2} + a_{3}x_{2}^{2} \\ f(x_{3}) = a_{0} + a_{1}x_{3} + a_{2}x_{3}^{2} + a_{3}x_{3}^{2} \\ \end{matrix} \right\}\ \ \ \ (4)$

Equation (4) can be expressed in matrix notation as

$\begin{bmatrix} 1 & x_{0} & x_{0}^{2} & x_{0}^{3} \\ 1 & x_{1} & x_{1}^{2} & x_{1}^{3} \\ 1 & x_{2} & x_{2}^{2} & x_{2}^{3} \\ 1 & x_{3} & x_{3}^{2} & x_{3}^{3} \\ \end{bmatrix}\begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \\ \end{bmatrix} = \begin{bmatrix} f\left( x_{0} \right) \\ f\left( x_{1} \right) \\ f\left( x_{2} \right) \\ f\left( x_{3} \right) \\ \end{bmatrix}\ \ \ (5)$

The above Equation (5) can symbolically be represented as

$\left\lbrack A \right\rbrack_{4 \times 4}{\overrightarrow{a}}_{4 \times 1} = {\overrightarrow{f}}_{4 \times 1}\ \ \ (6)$

Thus,

$\overrightarrow{a} = \begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \\ \end{bmatrix} = \left\lbrack A \right\rbrack^{- 1} \times \overrightarrow{f}\ \ \ (7)$

Substituting Equation (7) into Equation (3), one gets

$f_{3}\left( x \right) = \left\{ 1,x,x^{2},x^{3} \right\} \times \left\lbrack A \right\rbrack^{- 1} \times \overrightarrow{f}\ \ \ (8)$

a)  As indicated in Figure 1, one has

b)

$\begin{array}{rl} x_{0} & =a \\ x_{1} & =a+h \\ & =a+\frac{b-a}{3} \\ & =\frac{2 a+b}{3} \\ x_{2} & =a+2 h \\ & =a+\frac{2 b-2 a}{3} \\ & =\frac{a+2 b}{3} \\ x_{3} & =a+3 h \\ & =a+\frac{3 b-3 a}{3} \\ &= b \end{array} \ \ \ (9)$

With the help from MATLAB $$Ref. 2$$, the unknown vector $$\overrightarrow{a}$$ (shown in Equation 7) can be solved for symbolically.

Method 2

Using Lagrange interpolation, the cubic polynomial function $$f_{3}\left( x \right)$$ that passes through 4 data points (see Figure 1) can be explicitly given as

$\begin{split} f_{3}\left( x \right) &= \frac{\left( x - x_{1} \right)\left( x - x_{2} \right)\left( x - x_{3} \right)}{\left( x_{0} - x_{1} \right)\left( x_{0} - x_{2} \right)\left( x_{0} - x_{3} \right)} \times f\left( x_{0} \right) + \frac{\left( x - x_{0} \right)\left( x - x_{2} \right)\left( x - x_{3} \right)}{\left( x_{1} - x_{0} \right)\left( x_{1} - x_{2} \right)\left( x_{1} - x_{3} \right)} \times f\left( x_{1} \right)\\ & \ \ \ \ +\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)\left(x-x_{3}\right)}{\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)} \times f\left(x_{3}\right)+\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)\left(x-x_{2}\right)}{\left(x_{3}-x_{0}\right)\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)} \times f\left(x_{3}\right)\ \ \ (10) \end{split}$

## Simpsons 3/8 Rule for Integration

Substituting the form of $$\mathbf{{f}_{{3}}\left({x} \right)}$$ from Method (1) or Method (2),

$\begin{split} I &= \int_{a}^{b}{f\left( x \right){dx}}\\ &= \int_{a}^{b}{f_{3}\left( x \right){dx}}\\ &= \left( b - a \right) \times \frac{\left\{ f\left( x_{0} \right) + 3f\left( x_{1} \right) + 3f\left( x_{2} \right) + f\left( x_{3} \right) \right\}}{8}\ \ \ (11) \end{split}$

Since

$h = \frac{b - a}{3}$

$b - a = 3h$

and Equation (11) becomes

$I \approx \frac{3h}{8} \times \left\{ f\left( x_{0} \right) + 3f\left( x_{1} \right) + 3f\left( x_{2} \right) + f\left( x_{3} \right) \right\}\ \ \ (12)$

Note the 3/8 in the formula, and hence the name of method as the Simpson’s 3/8 rule.

The true error in Simpson 3/8 rule can be derived as $$Ref. 1$$

$E_{t} = - \frac{(b - a)^{5}}{6480} \times f^{\prime\prime\prime\prime}(\zeta) ,\ \text{where}\ a \leq \zeta \leq b\ \ \ (13)$

Example 1

The vertical distance in meters covered by a rocket from $$t = 8$$ to $$t = 30$$ seconds is given by

$s = \int_{8}^{30}{\left( 2000\ln\left\lbrack \frac{140000}{140000 - 2100t} \right\rbrack - 9.8t \right){dt}}$

Use Simpson 3/8 rule to find the approximate value of the integral.

Solution

$\begin{split} h &= \frac{b - a}{n}\\ &=\frac{b-a}{3}\\ &= \frac{30-8}{3}\\ &= 7.3333\end{split}$

$f(t) = 2000\ln\left\lbrack \frac{140000}{140000 - 2100t} \right\rbrack - 9.8t$

$I \approx \frac{3h}{8} \times \left\{ f\left( t_{0} \right) + 3f\left( t_{1} \right) + 3f\left( t_{2} \right) + f\left( t_{3} \right) \right\}$

$t_{0} = 8$

$\begin{split} f\left( t_{0} \right) &= 2000\ln\left( \frac{140000}{140000 - 2100 \times 8} \right) - 9.8 \times 8 \\ &= 177.2667 \end{split}$

$\begin{split} t_{1} &= t_{0} + h \\ &= 8 + 7.3333 \\ &= 15.3333 \end{split}$ $\begin{split} f\left( t_{1} \right) &= 2000\ln\left( \frac{140000}{140000 - 2100 \times 15.3333} \right) - 9.8 \times 15.3333 \\ &= 372.4629\end{split}$

$\begin{split} t_{2} &= t_{0} + 2h \\ &= 8 + 2(7.3333) \\ &= 22.6666 \end{split}$ $\begin{split} f\left( t_{2} \right) &= 2000\ln\left( \frac{140000}{140000 - 2100 \times 22.6666} \right) - 9.8 \times 22.6666 \\ &= 608.8976 \end{split}$

$\begin{split} t_{3} &= t_{0} + 3h \\ &= 8 + 3(7.3333) \\ &= 30 \end{split}$ $\begin{split} f\left( t_{3} \right) &= 2000\ln\left( \frac{140000}{140000 - 2100 \times 30} \right) - 9.8 \times 30 \\ &= 901.6740\end{split}$

Applying Equation (12), one has

$\begin{split} I &= \frac{3}{8} \times 7.3333 \times \left\{ 177.2667 + 3 \times 372.4629 + 3 \times 608.8976 + 901.6740 \right\}\\ &=11063.3104m \end{split}$

The exact answer can be computed as

$I_{\text{exact}} = 11061.34m$

## Composite Simpson 3/8 Rule

Using $$n$$ = number of equal segments, the width $$h$$can be defined as

$h = \frac{b - a}{n}\ \ \ (14)$

The number of segments need to be an integer multiple of 3 as a single application of Simpson 3/8 rule requires 3 segments.

The integral shown in Equation (1) can be expressed as

$\begin{split} I &= \int_{a}^{b}{f\left( x \right){dx}}\\ &\approx \int_{a}^{b}{f_{3}\left( x \right){dx}}\\ &\approx \int_{x_{0} = a}^{x_{3}}{f_{3}\left( x \right){dx}} + \int_{x_{3}}^{x_{6}}{f_{3}\left( x \right){dx}} + ........ + \int_{x_{n - 3}}^{x_{n} = b}{f_{3}\left( x \right){dx}}\ \ \ (15) \end{split}$

Using Simpson 3/8 rule (See Equation 12) into Equation (15), one gets

$\begin{split} I &= \frac{3h}{8}\begin{Bmatrix} f\left( x_{0} \right) + 3f\left( x_{1} \right) + 3f\left( x_{2} \right) + f\left( x_{3} \right) + f\left( x_{3} \right) + 3f\left( x_{4} \right) + 3f\left( x_{5} \right) + f\left( x_{6} \right) \\ + ..... + f\left( x_{n - 3} \right) + 3f\left( x_{n - 2} \right) + 3f\left( x_{n - 1} \right) + f\left( x_{n} \right) \\ \end{Bmatrix}\ \ \ (16)\\ &= \frac{3h}{8}\left\{ f\left( x_{0} \right) + 3\sum_{i = 1,4,7,..}^{n - 2}{f\left( x_{i} \right)} + 3\sum_{i = 2,5,8,..}^{n - 1}{f\left( x_{i} \right)} + 2\sum_{i = 3,6,9,..}^{n - 3}{f\left( x_{i} \right)} + f\left( x_{n} \right) \right\}\ \ \ (17)\end{split}$

Example 2

The vertical distance in meters covered by a rocket from $$t = 8$$ to $$t = 30$$ seconds is given by

$s = \int_{8}^{30}{\left( 2000\ln\left\lbrack \frac{140000}{140000 - 2100t} \right\rbrack - 9.8t \right){dt}}$

Use composite Simpson 3/8 rule with six segments to estimate the vertical distance.

Solution

In this example, one has (see Equation 14):

$f(t) = 2000\ln\left\lbrack \frac{140000}{140000 - 2100t} \right\rbrack - 9.8t$

$h = \frac{30 - 8}{6} = 3.6666$

$\left\{ t_{0},f\left( t_{0} \right) \right\} = \left\{ 8,177.2667 \right\}$

$$\left\{ t_{1},f\left( t_{1} \right) \right\} = \left\{ 11.6666,270.4104 \right\}\ \text{where}\ t_{1} = t_{0} + h = 8 + 3.6666 = 11.6666$$

$\left\{ t_{2},f\left( t_{2} \right) \right\} = \left\{ 15.3333,372.4629 \right\}\ \text{where}\ t_{2} = t_{0} + 2h = 15.3333$

$\left\{ t_{3},f\left( t_{3} \right) \right\} = \left\{ 19,484.7455 \right\}\ \text{where}\ t_{3} = t_{0} + 3h = 19$

$\left\{ t_{4},f\left( t_{4} \right) \right\} = \left\{ 22.6666,608.8976 \right\}\ \text{where}\ t_{4} = t_{0} + 4h = 22.6666$

$\left\{ t_{5},f\left( t_{5} \right) \right\} = \left\{ 26.3333,746.9870 \right\}\ \text{where}\ t_{5} = t_{0} + 5h = 26.3333$

$\left\{ t_{6},f\left( t_{6} \right) \right\} = \left\{ 30901.6740 \right\}\ \text{where}\ t_{6} = t_{0} + 6h = 30$

Applying Equation (17), one obtains:

$\begin{split} I&= \frac{3}{8}\left( 3.6666 \right)\left\{ 177.2667 + 3\sum_{i = 1,4,..}^{n - 2 = 4}{f\left( t_{i} \right)} + 3\sum_{i = 2,5,..}^{n - 1 = 5}{f\left( t_{i} \right)} + 2\sum_{i = 3,6,..}^{n - 3 = 3}{f\left( t_{i} \right)} + 901.6740 \right\}\\ &= \left( 1.3750 \right)\begin{Bmatrix} 177.2667 + 3\left( 270.4104 + 608.8976 \right) \\ + 3\left( 372.4629 + 746.9870 \right) + 2\left( 484.7455 \right) + 901.6740 \\ \end{Bmatrix}\\ &= 11601.4696m \end{split}$

Example 3

Compute

$I = \int_{8}^{30}{\left\{ 2000\ln\left( \frac{140000}{140000 - 2100t} \right) - 9.8t \right\}{dt},}$

using Simpson 1/3 rule (with $$n_{1} =4$$), and Simpson 3/8 rule (with $$n_{2} =3$$).

Solution

The segment width is

$\begin{split} h &= \frac{b - a}{n}\\ &= \frac{b - a}{n_{1} + n_{2}}\\ &= \frac{30 - 8}{\left( 4 + 3 \right)}\\ &= 3.1429 \end{split}$

$f(t) = 2000\ln\left\lbrack \frac{140000}{140000 - 2100t} \right\rbrack - 9.8t$

$\begin{split} &{\left. \ \begin{matrix} t_{0} = a = 8 \\ t_{1} = x_{0} + 1h = 8 + 3.1429 = 11.1429 \\ t_{2} = t_{0} + 2h = 8 + 2\left( 3.1429 \right) = 14.2857 \\ t_{3} = t_{0} + 3h = 8 + 3\left( 3.1429 \right) = 17.4286 \\ t_{4} = t_{0} + 4h = 8 + 4\left( 3.1429 \right) = 20.5714 \\ \end{matrix} \right\}\text{Simpson's 1/3rule}}\\ &t_{5} = t_{0} + 5h = 8 + 5\left( 3.1429 \right) = 23.7143\\ &t_{6} = t_{0} + 6h = 8 + 6\left( 3.1429 \right) = 26.8571\\ &t_{7} = t_{0} + 7h = 8 + 7\left( 3.1429 \right) = 30 \end{split}$

Now

$\begin{split} f\left( t_{0} = 8 \right) &= 2000\ln\left( \frac{140,000}{140,000 - 2100 \times 8} \right) - 9.8 \times 8\\ &= 177.2667 \end{split}$

Similarly:

$f\left( t_{1} \right) = 256.5863$

$f\left( t_{2} \right) = 342.3241$

$f\left( t_{3} \right) = 435.2749$

$f\left( t_{4} \right) = 536.3909$

$f\left( t_{5} \right) = 646.8260$

$f\left( t_{6} \right) = 767.9978$

$f\left( t_{7} \right) = 901.6740$

For several segments $$\left( n_{1} = \text{first}\ 4\ \text{segments} \right)$$, using composite Simpson 1/3 rule, one obtains (See Equation 19):

$\begin{split} I_{1} &= \left( \frac{h}{3} \right)\left\{ f\left( t_{0} \right) + 4\sum_{i = 1,3,...}^{n_{1} - 1 = 3}{f\left( t_{i} \right)} + 2\sum_{i = 2,...}^{n_{1} - 2 = 2}{f\left( t_{i} \right)} + f\left( t_{n_{1}} \right) \right\}\\ &= \left( \frac{h}{3} \right)\left\{ f\left( t_{0} \right) + 4\left( f\left( t_{1} \right) + f\left( t_{3} \right) \right) + 2f\left( t_{2} \right) + f\left( t_{4} \right) \right\}\\ &= \left( \frac{3.1429}{3} \right)\left\{ 177.2667 + 4\left( 256.5863 + 435.2749 \right) + 2\left( 342.3241 \right) + 536.3909 \right\} \\ &= 4364.1197 \end{split}$

For several segments $$\left( n_{2} = \text{last 3 segments} \right)$$, using single application Simpson 3/8 rule, one obtains (See Equation 17):

$\begin{split} I_{2} &= \left( \frac{3h}{8} \right)\left\{ f\left( t_{0} \right) + 3\sum_{i = 1,3,...}^{n_{2} - 2 = 1}{f\left( t_{i} \right)} + 3\sum_{i = 2,...}^{n_{2} - 1 = 2}{f\left( t_{i} \right)} + 2\sum_{i = 3,6,..}^{n_{2} - 3 = 0}{f\left( t_{i} \right) +}f\left( t_{n_{1}} \right) \right\}\\ &= \left( \frac{3h}{8} \right)\left\{ f\left( t_{0} \right) + 3f\left( t_{1} \right) + 3f\left( t_{2} \right) + 2(\text{no contribution}) + f(t_{3}) \right\} \\ &= \left( \frac{3h}{8} \right)\left\{ f\left( t_{4} \right) + 3f\left( t_{5} \right) + 3f\left( t_{6} \right) + f(t_{7}) \right\}\\ &= \left( \frac{3}{8} \times 3.1429 \right)\left\{ 536.3909 + 3\left( 646.8260 \right) + 3\left( 767.9978 \right) + 901.6740 \right\}\\ &= 6697.3663 \end{split}$

The mixed (combined) Simpson 1/3 and 3/8 rules give

$\begin{split} I &= I_{1} + I_{2}\\ &= 4364.1197 + 6697.3663\\ &= 11061m \end{split}$

c)  Comparing the truncated error of Simpson 1/3 rule

d)  $E_{t} = - \frac{\left( b - a \right)^{5}}{2880} \times f^{\prime\prime\prime\prime}\zeta)\ \ \ (18)$

With Simpson 3/8 rule (See Equation 12), it seems to offer slightly more accurate answer than the former. However, the cost associated with Simpson 3/8 rule (using 3rd order polynomial function) is significantly higher than the one associated with Simpson 1/3 rule (using 2nd order polynomial function).

The number of segments that can be used in the conjunction with Simpson 1/3 rule is 2, 4, 6, 8, … (any even numbers) for

$\begin{split} I &= \int_{a}^{b}{f(x){dx}}\\ &\approx \left( \frac{h}{3} \right)\left\{ f\left( x_{0} \right) + 4f\left( x_{1} \right) + f\left( x_{2} \right) + f\left( x_{2} \right) + 4f\left( x_{3} \right) + f\left( x_{4} \right) + ... + f\left( x_{n - 2} \right) + 4f\left( x_{n - 1} \right) + f\left( x_{n} \right) \right\}\\&= \begin{array}{l} \quad\left(\displaystyle \frac{h}{3}\right)\left\{f\left(x_{0}\right)+4 \displaystyle\sum_{i=1,3, ...}^{n-1} f\left(x_{i}\right)+2 \sum_{i=2,4,6 ...}^{n-2} f\left(x_{i}\right)+\right. \left.f\left(x_{n}\right)\right\} \end{array}\end{split}$

However, Simpson 3/8 rule can be used with the number of segments equal to 3,6,9,12,.. (can be certain integers that are multiples of 3).

If the user wishes to use, say 7 segments, then the mixed Simpson 1/3 rule (for the first 4 segments), and Simpson 3/8 rule (for the last 3 segments) would be appropriate.

## Computer Algorithm for Mixed Simpson 1/3 and 3/8 Rule for Integration

Based on the earlier discussion on (single and composite) Simpson 1/3 and 3/8 rules, the following “pseudo” step-by-step mixed Simpson rules for estimating

$I = \int_{a}^{b}{f(x){dx}}$

can be given as

Step 1

User inputs information, such as

$f(x) = \text{integrand}$

$n_{1} = \text{number of segments in conjunction with Simpson 1/3 rule (a multiple of 2 (any even numbers)}$

$n_{2}= \text{number of segments in conjunction with Simpson 3/8 rule (a multiple of 3)}$

Step 2

Compute

$n = n_{1} + n_{2}$

$h = \frac{b - a}{n}$

$x_{0} = a$

$x_{1} = a + 1h$

$x_{2} = a + 2h$ $\vdots$

$x_{i} = a + ih$ $\vdots$ $x_{n} = a + nh = b$

Step 3

Compute result from composite Simpson 1/3 rule (See Equation 19)

$I_{1} = \left( \frac{h}{3} \right)\left\{ f\left( x_{0} \right) + 4\sum_{i = 1,3,...}^{n_{1} - 1}{f\left( x_{i} \right)} + 2\sum_{i = 2,4,6...}^{n_{1} - 2}{f\left( x_{i} \right)} + f\left( x_{n_{1}} \right) \right\}\ \ \ (19, repeated)$

Step 4

Compute result from composite Simpson 3/8 rule (See Equation 17)

$\begin{split} I_{2} = \left( \frac{3h}{8} \right)\left\{ f\left( x_{0} \right) + 3\sum_{i = 1,4,7...}^{n_{2} - 2}{f\left( x_{i} \right)} + 3\sum_{i = 2,5,8...}^{n_{2} - 1}{f\left( x_{i} \right)} + 2\sum_{i = 3,6,9,...}^{n_{2} - 3}{f\left( x_{i} \right) +}f\left( x_{n_{2}} \right) \right\}\ \ \ (17, repeated)\end{split}$

Step 5

$I \approx I_{1} + I_{2}\ \ \ (20)$

and print out the final approximated answer for $$I$$.

## Multiple Choice Test

(1). Simpson 3/8 rule for integration is mainly based upon the idea of

(A) approximating $$\displaystyle {f(x)}$$ in $$\displaystyle I = \int_{a}^{b}{f(x)dx}$$ by a cubic polynomial

(B) approximating $$\displaystyle {f(x)}$$ in $$\displaystyle I = \int_{a}^{b}{f(x)dx}$$ by a quadratic polynomial

(C) Converting the limit of integral limits $$\lbrack a,b\rbrack$$ into $$\lbrack - 1, + 1\rbrack$$

(D) Using similar concepts as Gauss quadrature formula

(2). The exact value of $$\displaystyle \int_{1}^{4}{(e^{- 2x} + 4x^{2} - 8)dx}$$ most nearly is

(A) $$6.0067$$

(B) $$5.7606$$

(C) $$60.0675$$

(D) $$67.6075$$

(3). The approximate value of $$\displaystyle \int_{1}^{4}{(e^{- 2x} + 4x^{2} - 8)dx}$$ by a single application of Simpson’s 3/8 rule is

(A) $$61.3740$$

(B) $$60.0743$$

(C) $$59.3470$$

(D) $$58.8992$$

(4). The approximate value of $$\displaystyle \int_{1}^{4}{(e^{- 2x} + 4x^{2} - 8)dx}$$ by a composite Simpson’s 3/8 rule with n=6 segments is most nearly

(A) $$60.8206$$

(B) $$60.6028$$

(C) $$61.0677$$

(D) $$60.0675$$

(5). The approximate value of $$\displaystyle \int_{1}^{4}{(e^{- 2x} + 4x^{2} - 8)dx}$$ by combination of Simpson’s 1/3 rule (n=6 segments) and Simpson’s 3/8 rule (n=3 segments) most nearly is

(A) $$60.0677$$

(B) $$59.0677$$

(C) $$61.0677$$

(D) $$59.7607$$

(6). Comparing Simpson’s 3/8 rule truncated error formula

$E_{t} = - \displaystyle \frac{(b - a)^{5}}{6480} \times f^{(4)}\left( \zeta \right),\ a \leq \zeta \leq b,$

with Simpson’s 1/3 rule truncated error formula

$E_{t}=-\displaystyle \frac{(b-a)^{5}}{2880} f^{(4)}(\zeta),\ \quad a<\zeta<b$

the following conclusion can be made.

(A) Simpson’s 3/8 rule is significantly more accurate than Simpson’s 1/3 rule

(B) It is worth it in terms of computational efforts versus accuracy to use Simpson’s 3/8 rule instead of Simpson’s 1/3 rule.

(C) It is worth it in terms of computational efforts versus accuracy to use Simpson’s 3/8 rule instead of Simpson’s 1/3 rule.

(D) Simpson’s 3/8 rule is less accurate than Simpson’s 1/3 rule.

## Problem Set

(1). Compute the following integral exactly $\int_{1}^{4}{x^{2}e^{(3x)}dx}$ Answer: $$7.3541 \times 10^5$$

/ /

(2). Using composite Simpson 3/8 rule (with $$n = 6$$ “little” segments), compute the integral

$\int_{1}^{4}{x^{2}e^{(3x)}dx}$ Answer: $$I = 784785.6554$$

/ /

(3). Using the MIXED composite Simpson 1/3 rule (with $$n_{1} = 4$$ “little” segments), and Simpson 3/8 rule (with $$n_{2} = 6$$ “little” segments), compute the integral $$\displaystyle \int_{1}^{4}{x^{2}e^{(3x)}dx}$$

Answer: $$I = 744173.7172$$

/ /

(4). Compute the following integral exactly $\int_{\pi}^{2\pi}{x\sin(2x)}dx$

Answer: $$I = -1.5708$$

/ /

(5). Using the single application Simpson 3/8 rule (with $$n_{2} = 3$$ “little” segments), compute the integral $$\displaystyle \int_{\pi}^{2\pi}{x\sin(2x)}dx$$

Answer: $$I = -1.0684$$

/ /

(6). Using the composite Simpson 3/8 rule (with $$n_{2} = 6$$ “little” segments), compute the integral $$\displaystyle \int_{\pi}^{2\pi}{x\sin(2x)}dx$$

Answer: $$I = -1.6026$$

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(7). Using the MIXED, single-application Simpson 1/3 and 3/8 rules (with $$n_{1} = 2$$, and $$n_{2} = 3$$ “little” segments), compute the integral $$\displaystyle \int_{\pi}^{2\pi}{x\sin(2x)}dx$$

Answer: $$I = -1.7357$$