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: \(7.3541 \times 10^5\)
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(2). Using composite Simpson 3/8 rule (with \(n = 6\) “little” segments), compute the integral
\[\int_{1}^{4}{x^{2}e^{(3x)}dx}\] Answer: \(I = 784785.6554\)
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(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: \(I = 744173.7172\)
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(4). Compute the following integral exactly \[\int_{\pi}^{2\pi}{x\sin(2x)}dx\]
Answer: \(I = -1.5708\)
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(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: \(I = -1.0684\)
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(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: \(I = -1.7357\)