Chapter 07.03: Simpson’s 1/3rd Rule of Integration
Learning Objective
After successful completion of this lesson, you should be able to:
1) derive the single application of Simpson’s 1/3rd rule formula,
2) use the single application of Simpson’s 1/3rd rule of integration to solve problems.
What is integration?
Integration is the process of measuring the area under a function plotted on a graph. Why would we want to integrate a function? Among the most common examples are finding the velocity of a body from an acceleration function, and displacement of a body from a velocity function. Throughout many engineering fields, there are (what sometimes seems like) countless applications for integral calculus. You can read about some of these applications in Chapters 07.00A-07.00G.
Sometimes, the evaluation of expressions involving these integrals can become daunting, if not indeterminate. For this reason, a wide variety of numerical methods has been developed to simplify the integral. Here, we will discuss Simpson’s 1/3 rule of integral approximation, which improves upon the accuracy of the trapezoidal rule.
Here, we will discuss the Simpson’s 1/3 rule of approximating integrals of the form
\[I = \int_{a}^{b}{f\left( x \right){dx}}\]
where
\[f(x)\ \text{is called the integrand,}\]
\[a = \text{lower limit of integration}\]
\[b = \text{upper limit of integration}\]
Simpson’s 1/3 Rule
The trapezoidal rule was based on approximating the integrand by a first order polynomial, and then integrating the polynomial over interval of integration. Simpson’s 1/3 rule is an extension of Trapezoidal rule where the integrand is approximated by a second order polynomial.
Figure 1 Integration of a function
Hence
\[I= \int_{a}^{b}{f\left( x \right){dx}} \approx \int_{a}^{b}{f_{2}\left( x \right){dx}}\]
where \(f_{2}(x)\) is a second order polynomial given by
\[f_{2}(x) = a_{0} + a_{1}x + a_{2}x^2\]
Choose
\[(a,f(a)),\ \left( \frac{a + b}{2},f\left( \frac{a + b}{2} \right) \right),\ \text{and}\ (b,f(b))\]
as the three points of the function to evaluate \(a_{1}\) and \(a_{2}\).
\[f(a)=f_{2}(a)=a_{0}+a_{1} a+a_{2} a^{2}\]
\[f\left(\frac{a+b}{2}\right)=f_{2}\left(\frac{a+b}{2}\right)=a_{0}+a_{1}\left(\frac{a+b}{2}\right)+a_{2}\left(\frac{a+b}{2}\right)^{2}\]
\[f(b)=f_{2}(b)=a_{0}+a_{1} b+a_{2} b^{2}\]
Solving the above three equations for unknowns, \(a_{0},\) \(a_{1}\) and \(a_{2}\) give
\[a_{0}=\frac{a^{2} f(b)+a b f(b)-4 a b f\left(\frac{a+b}{2}\right)+a b f(a)+b^{2} f(a)}{a^{2}-2 a b+b^{2}}\]
\[a_{1}=-\frac{a f(a)-4 a f\left(\frac{a+b}{2}\right)+3 a f(b)+3 b f(a)-4 b f\left(\frac{a+b}{2}\right)+b f(b)}{a^{2}-2 a b+b^{2}}\]
\[a_{2}=\frac{2\left(f(a)-2 f\left(\frac{a+b}{2}\right)+f(b)\right)}{a^{2}-2 a b+b^{2}}\]
Then
\[\begin{split} I & \approx \int_{a}^{b} f_{2}(x) d x \\ &=\int_{a}^{b}\left(a_{0}+a_{1} x+a_{2} x^{2}\right) d x \\ &=\left[a_{0} x+a_{1} \frac{x^{2}}{2}+a_{2} \frac{x^{3}}{3}\right]_{a}^{b} \\ &=a_{0}(b-a)+a_{1} \frac{b^{2}-a^{2}}{2}+a_{2} \frac{b^{3}-a^{3}}{3} \end{split}\]
Substituting values of \(a_{0},\) \(a_{1}\) and \(a_{2}\) give
\[\int_{a}^{b} f_{2}(x) d x=\frac{b-a}{6}\left[f(a)+4 f\left(\frac{a+b}{2}\right)+f(b)\right]\]
Since for the single application of Simpson 1/3 rule, the interval \(\left\lbrack a,b \right\rbrack\) is broken into 2 segments, the segment width
\[h = \frac{b-a}{2}\]
The Simpson’s 1/3 rule can be rewritten as
\[\int_{a}^{b}{f(x)dx} \approx \frac{h}{3}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack\]
Since the above form has 1/3 in its formula, it is called Simpson’s 1/3 rule.
Example 1
The distance covered by a rocket in meters from \(t = 8\)s to \(t = 30\)s is given by
\[x=\int_{8}^{30}\left(2000 \ln \left[\frac{140000}{140000-2100 t}\right]-9.8 t\right) dt\]
a) Use Simpson’s 1/3 rule to find the approximate value of \(x\).
b) Find the true error, \(E_{t}\).
c) Find the absolute relative true error, \(\left| \epsilon_{t} \right|\).
Solution
a)
\[x \approx \frac{b - a}{6}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack\]
\[a=8\]
\[b=30\]
\[\frac{a+b}{2}=19\]
\[f(t)=2000 \ln \left[\frac{140000}{140000-2100 t}\right]-9.8 t\]
\[f(8)=2000 \ln \left[\frac{140000}{140000-2100(8)}\right]-9.8(8)=177.27 {m} / {s}\]
\[f(30)=2000 \ln \left[\frac{140000}{140000-2100(30)}\right]-9.8(30)=901.67 {m} / {s}\]
\[f(19)=2000 \ln \left(\frac{140000}{140000-2100(19)}\right)-9.8(19)=484.75 {m} / {s}\]
\[\begin{split} x & \approx \frac{b-a}{6}\left[f(a)+4 f\left(\frac{a+b}{2}\right)+f(b)\right] \\ &=\left(\frac{30-8}{6}\right)[f(8)+4 f(19)+f(30)] \\ &=\frac{22}{6}[177.27+4 \times 484.75+901.67] \\ &=11065.72 {m} \end{split}\]
b) The exact value of the above integral is
\[\begin{split} x &=\int_{8}^{30}\left(2000 \ln \left[\frac{140000}{140000-2100 t}\right]-9.8 t\right) d t \\ &=11061.34 {m} \end{split}\]
So the true error is
\[\begin{split} E_{t} &=\text { True Value}-\text{Approximate Value } \\ &=11061.34-11065.72 \\ &=-4.38 m \end{split}\]
c) The absolute relative true error is
\[\begin{split}\left|\epsilon_{t}\right| &=\left|\frac{\text { True Error }}{\text { True Value }}\right| \times 100 \\ &=\left|\frac{-4.38}{11061.34}\right| \times 100 \\ &=0.0396 \% \end{split}\]
Appendix: Alternative Methods of Deriving Simpson’s 1/3rd rule (Not in syllabus)
Alternative Method 1:
Simpson’s 1/3 rule can also be derived by approximating \(f(x)\) by a second order polynomial using Newton’s divided difference polynomial as
\[f_{2}(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)\left(x-\frac{a+b}{2}\right)\]
where
\[b_{0}=f(a)\]
\[b_{1}=\frac{f\left(\displaystyle \frac{a+b}{2}\right)-f(a)}{\displaystyle \frac{a+b}{2}-a}\]
\[b_{2}=\frac{\displaystyle \frac{f(b)-f\left(\displaystyle \frac{a+b}{2}\right)}{\displaystyle b-\frac{a+b}{2}}-\frac{\displaystyle f\left(\displaystyle \frac{a+b}{2}\right)-f(a)}{\displaystyle \frac{a+b}{2}-a}}{\displaystyle b-a}\]
Integrating Newton’s divided difference polynomial gives us
\[\begin{split} \int_{a}^{b} f(x) d x &\approx \int_{a}^{b} f_{2}(x) d x \\ &=\int_{a}^{b}\left[b_{0}+b_{1}(x-a)+b_{2}(x-a)\left(x-\frac{a+b}{2}\right)\right] dx \\ &=\left[b_{0} x+b_{1}\left(\frac{x^{2}}{2}-a x\right)+b_{2}\left(\frac{x^{3}}{3}-\frac{(3 a+b) x^{2}}{4}+\frac{a(a+b) x}{2}\right)\right]_{a}^{b}\\ &=b_{0}(b-a)+b_{1}\left(\frac{b^{2}-a^{2}}{2}-a(b-a)\right)\\ & \ \ \ +b_{2}\left(\frac{b^{3}-a^{3}}{3}-\frac{(3a+b)\left(b^{2}-a^{2}\right)}{4}+\frac{a(a+b)(b-a)}{2}\right)\end{split}\]
Substituting values of \(b_{0},\) \(b_{1},\) and \(b_{2}\) into this equation yields the same result as before
\[\begin{split} \int_{a}^{b}{f(x)dx} &\approx \frac{b - a}{6}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack\\ &= \frac{h}{3}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack \end{split}\]
Alternative Method 2:
One could even use the Lagrange polynomial to derive Simpson’s formula. Notice any method of three-point quadratic interpolation can be used to accomplish this task. In this case, the interpolating function becomes
\[\begin{split} f_{2}(x) &=\frac{\left(x-\displaystyle \frac{a+b}{2}\right)(x-b)}{\left(a-\displaystyle \frac{a+b}{2}\right)(a-b)} f(a)+\displaystyle \frac{(x-a)(x-b)}{\left(\displaystyle \frac{a+b}{2}-a\right)\left(\displaystyle \frac{a+b}{2}-b\right)} f\left(\displaystyle \frac{a+b}{2}\right)\\ &\ \ \ \ \ \ +\displaystyle \frac{(x-a)\left(x-\displaystyle \frac{a+b}{2}\right)}{(b-a)\left(b-\displaystyle \frac{a+b}{2}\right)} f(b) \end{split}\]
Integrating this function gets
\[\begin{split} \int_{a}^{b} f_{2}(x) d x &=\left[\begin{array}{l}\displaystyle \frac{\displaystyle \frac{x^{3}}{3}-\displaystyle \frac{(a+3 b) x^{2}}{4}+\displaystyle \frac{b(a+b) x}{2}}{\left(a-\displaystyle \frac{a+b}{2}\right)(a-b)} f(a)\\ +\displaystyle \frac{\displaystyle \frac{x^{3}}{3}-\displaystyle \frac{(a+b) x^{2}}{2}+a b x}{\left(\displaystyle \frac{a+b}{2}-a\right)\left(\displaystyle \frac{a+b}{2}-b\right)} f\left(\displaystyle \frac{a+b}{2}\right) \\ +\displaystyle \frac{\displaystyle \frac{x^{3}}{3}-\displaystyle \frac{(3 a+b) x^{2}}{4}+\displaystyle \frac{a(a+b) x}{2}}{(b-a)\left(b-\displaystyle \frac{a+b}{2}\right)} f(b)\end{array}\right]_{a}^{b}\\ &=\displaystyle \frac{\displaystyle \frac{b^{3}-a^{3}}{3}-\displaystyle \frac{(a+3 b)\left(b^{2}-a^{2}\right)}{4}+\displaystyle \frac{b(a+b)(b-a)}{2}}{\left(a-\displaystyle \frac{a+b}{2}\right)(a-b)} f(a)\\ & \ \ \ +\displaystyle \frac{\displaystyle \frac{b^{3}-a^{3}}{3}-\displaystyle \frac{(a+b)\left(b^{2}-a^{2}\right)}{2}+a b(b-a)}{\left(\displaystyle \frac{a+b}{2}-a\right)\left(\displaystyle \frac{a+b}{2}-b\right)} f\left(\displaystyle \frac{a+b}{2}\right)\\ & \ \ \ +\displaystyle \frac{\displaystyle \frac{b^{3}-a^{3}}{3}-\displaystyle \frac{(3 a+b)\left(b^{2}-a^{2}\right)}{4}+\displaystyle \frac{a(a+b)(b-a)}{2}}{(b-a)\left(b-\displaystyle \frac{a+b}{2}\right)} f(b) \end{split}\]
Believe it or not, simplifying and factoring this large expression yields you the same result as before
\[\begin{split} \int_{a}^{b}{f(x)dx} &\approx \frac{b - a}{6}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack\\ &= \frac{h}{3}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack \end{split}\]
Alternative Method 3:
Simpson’s 1/3 rule can also be derived by the method of coefficients. Assume
\[\int_{a}^{b}{f(x)dx} \approx c_{1}f(a) + c_{2}f\left( \frac{a + b}{2} \right) + c_{3}f(b)\]
Let the right-hand side be an exact expression for the integrals \(\displaystyle \int_{a}^{b}{1dx},\) \(\displaystyle \int_{a}^{b}{xdx},\) and \(\displaystyle \int_{a}^{b}{x^{2}{dx}}\). This implies that the right-hand side will be exact expressions for integrals of any linear combination of the three integrals for a general second order polynomial. Now
\[\int_{a}^{b} 1 d x=b-a=c_{1}+c_{2}+c_{3}\]
\[\int_{a}^{b} x d x=\frac{b^{2}-a^{2}}{2}=c_{1} a+c_{2} \frac{a+b}{2}+c_{3} b\]
\[\int_{a}^{b} x^{2} d x=\frac{b^{3}-a^{3}}{3}=c_{1} a^{2}+c_{2}\left(\frac{a+b}{2}\right)^{2}+c_{3} b^{2}\]
Solving the above three equations for \(c_{1},\) \(c_{2}\) and \(c_{3}\) give
\[c_{1}=\frac{b-a}{6}\]
\[c_{2}=\frac{2(b-a)}{3}\]
\[c_{3}=\frac{b-a}{6}\]
This gives
\[\begin{split} \int_{a}^{b}{f(x)dx} &\approx \frac{b - a}{6}f(a) + \frac{2(b - a)}{3}f\left( \frac{a + b}{2} \right) + \frac{b - a}{6}f(b)\\ &= \frac{b - a}{6}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack\\ &= \frac{h}{3}\left\lbrack f(a) + 4f\left( \frac{a + b}{2} \right) + f(b) \right\rbrack \end{split}\]
The integral from the first method
\[\int_{a}^{b}{f(x)dx} \approx \ \int_{a}^{b}{(a_{0} + a_{1}x + a_{2}x^{2})dx}\]
can be viewed as the area under the second order polynomial, while the equation from Method 4
\[\int_{a}^{b}{f(x)dx} \approx \frac{b - a}{6}f(a) + \frac{2(b - a)}{3}f\left( \frac{a + b}{2} \right) + \frac{b - a}{6}f(b)\]
can be viewed as the sum of the areas of three rectangles.
Learning Objective
After successful completion of this lesson, you should be able to:
1) derive the composite (also called multiple-segment) Simpson’s 1/3rd rule formula,
2) use the composite (also called multiple-segment) Simpson’s 1/3rd rule of integration to solve problems.
Composite Simpson’s 1/3 Rule
Just like in composite trapezoidal rule, one can subdivide the interval \(\left\lbrack a,b \right\rbrack\) into \(n\) segments and apply Simpson’s 1/3 rule repeatedly over every two segments. Note that \(n\) needs to be even. Divide interval \(\left\lbrack a,b \right\rbrack\) into \(n\) equal segments, so that the segment width is given by
\[h = \frac{b - a}{n}\]
Now
\[\int_{a}^{b} f(x) d x=\int_{x_{0}}^{x_{n}} f(x) dx\]
where
\[x_{0}=a\]
\[x_{n}=b\]
\[\int_{a}^{b} f(x) d x=\int_{x_{0}}^{x_{2}} f(x) d x+\int_{x_{2}}^{x_{4}} f(x) d x+\ldots \ldots+ \int_{x_{n-4}}^{x_{n-2}} f(x) d x+\int_{x_{n-2}}^{x_{n}} f(x) dx\]
Apply Simpson’s 1/3rd Rule over each interval,
\[\begin{split} \int_{a}^{b} f(x) d x &\cong\left(x_{2}-x_{0}\right)\left[\frac{f\left(x_{0}\right)+4 f\left(x_{1}\right)+f\left(x_{2}\right)}{6}\right]+\\ &\ \ \ \ \ \left(x_{4}-x_{2}\right)\left[\frac{f\left(x_{2}\right)+4 f\left(x_{3}\right)+f\left(x_{4}\right)}{6}\right]+\ldots+\\ & \ \ \ \ \ \left(x_{n-2}-x_{n-4}\right)\left[\frac{f\left(x_{n-4}\right)+4 f\left(x_{n-3}\right)+f\left(x_{n-2}\right)}{6}\right]+\\ &\ \ \ \ \ \left(x_{n}-x_{n-2}\right)\left[\frac{f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)}{6}\right] \end{split}\]
Since
\[x_{i}-x_{i-2}=2 h\]
\[i=2,4, \ldots, n\]
then
\[\int_{a}^{b} f(x) d x \cong 2 h\left[\frac{f\left(x_{0}\right)+4 f\left(x_{1}\right)+f\left(x_{2}\right)}{6}\right]+2 h\left[\frac{f\left(x_{2}\right)+4 f\left(x_{3}\right)+f\left(x_{4}\right)}{6}\right]+\ldots\] \[+2 h\left[\frac{f\left(x_{n-4}\right)+4 f\left(x_{n-3}\right)+f\left(x_{n-2}\right)}{6}\right]+2 h\left[\frac{f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)}{6}\right]\] \[= \frac{h}{3}\left[f(x_0)+4(f(x_1)+f(x_3)+...+f(x_{n-1}))+2(f(x_2)+f(x_4)+....+f(x_{n-2}))+f(x_n)\right]\] \[=\frac{h}{3}\left[f\left(x_{0}\right)+4 \sum_{\substack{i=1 \\ i=\text{odd}}}^{n-1} f\left(x_{i}\right)+2 \sum_{\substack{i=2\\ i=\text {even}}}^{n-2} f\left(x_{i}\right)+f\left(x_{n}\right)\right]\]
\[\int_{a}^{b} f(x) d x \cong \frac{b-a}{3 n}\left[f\left(x_{0}\right)+4 \sum_{\substack{i=1\\ \text{i=odd}}}^{n-1} f\left(x_{i}\right)+2 \sum_{\substack{i=2\\ i=\text{even}}}^{n-2} f\left(x_{i}\right)+f\left(x_{n}\right)\right]\]
Example 1
Use composite Simpson’s 1/3 rule with \(4\) segments to approximate the distance covered by a rocket in meters from \(t = 8\)s to \(t = 30\)s as given by
\[x=\int_{8}^{30}\left(2000 \ln \left[\frac{140000}{140000-2100 t}\right]-9.8 t\right) d t\]
a) Use composite Simpson’s 1/3rd rule with \(4\) segments to estimate x.
b) Find the true error, \(E_{t}\) for part (a).
c) Find the absolute relative true error, \(\left| \epsilon_{t} \right|\) for part (a).
Solution:
a) Using composite Simpson’s 1/3 rule,
\[x \approx \frac{b-a}{3 n}\left[f\left(t_{0}\right)+4 \sum_{\substack{i=1\\ \text{i=odd}}}^{n-1} f\left(t_{i}\right)+2 \sum_{\substack{i=2\\ i=\text{even}}}^{n-2} f\left(t_{i}\right)+f\left(t_{n}\right)\right]\]
\[n =4\]
\[a =8\]
\[b =30\]
\[\begin{split}h &=\frac{b-a}{n} \\ &=\frac{30-8}{4} \\ &=5.5 \\ f(t) &=2000 \ln \left[\frac{140000}{140000-2100 t}\right]-9.8 t \end{split}\]
So
\[f\left(t_{0}\right)=f(8)\]
\[\begin{split} f(8)&=2000 \ln \left[\frac{140000}{140000-2100(8)}\right]-9.8(8)\\ &=177.27 {m} / {s}\end{split}\]
\[f\left(t_{1}\right)=f(8+5.5)=f(13.5)\]
\[\begin{split} f(13.5)&=2000 \ln \left[\frac{140000}{140000-2100(13.5)}\right]-9.8(13.5)\\ &=320.25 {m} / {s}\end{split}\]
\[f\left(t_{2}\right)=f(13.5+5.5)=f(19)\]
\[\begin{split} f(19) &=2000 \ln \left(\frac{140000}{140000-2100(19)}\right)-9.8(19)\\ &=484.75 {m} / {s} \end{split}\]
\[f\left(t_{3}\right)=f(19+5.5)=f(24.5)\]
\[\begin{split} f(24.5)&=2000 \ln \left[\frac{140000}{140000-2100(24.5)}\right]-9.8(24.5)\\ &=676.05 {m} / {s}\end{split}\]
\[f\left(t_{4}\right)=f\left(t_{n}\right)=f(30)\]
\[\begin{split} f(30)&=2000 \ln \left[\frac{140000}{140000-2100(30)}\right]-9.8(30)\\ &=901.67 {m} / {s} \end{split}\]
\[\begin{split} x &=\frac{b-a}{3 n}\left[f\left(t_{0}\right)+4 \sum_{\substack{i=1\\ i=\text{odd}}}^{n-1} f\left(t_{i}\right)+2 \sum_{\substack{i=2\\ i= \text{even}}}^{n-2} f\left(t_{i}\right)+f\left(t_{n}\right)\right] \\ &=\frac{30-8}{3(4)}\left[f(8)+4 \sum_{\substack{i=1 \\ i=\text{odd}}}^{3} f\left(t_{i}\right)+2 \sum_{\substack{i=2 \\ i=\text {even }}}^{2} f\left(t_{i}\right)+f(30)\right] \\ &=\frac{22}{12}\left[f(8)+4 f\left(t_{1}\right)+4 f\left(t_{3}\right)+2 f\left(t_{2}\right)+f(30)\right] \\ &=\frac{11}{6}[f(8)+4 f(13.5)+4 f(24.5)+2 f(19)+f(30)] \\ &=\frac{11}{6}[177.27+4(320.25)+4(676.05)+2(484.75)+901.67] \\ &=11061.64 {m} \end{split}\]
b) The exact value of the above integral is
\[\begin{split} x &=\int_{8}^{30}\left(2000 \ln \left[\frac{140000}{140000-2100 t}\right]-9.8 t\right) d t \\ &=11061.34 {m} \end{split}\]
So the true error is
\[\begin{split} E_{t} &=\text { True Value} - \text{Approximate Value } \\ E_{t} &=11061.34-11061.64 \\ &=-0.30 {m} \end{split}\]
c) The absolute relative true error is
\[\begin{split}\left|\epsilon_{t}\right| &=\left|\frac{\text { True Error }}{\text { True Value }}\right| \times 100 \\ &=\left|\frac{-0.3}{11061.34}\right| \times 100 \\ &=0.0027 \% \end{split}\]
Table 1 Values of composite Simpson’s 1/3 rule for Example 1
| \(\text{n}\) | \(\text{Approximate Value}\) | \({E_{t}}\) | \({|\varepsilon_{t}|}\) |
|---|---|---|---|
\(2\) \(4\) \(6\) \(8\) \(10\) |
\(11065.72\) \(11061.64\) \(11061.40\) \(11061.35\) \(11061.34\) |
\(-4.38\) \(-0.30\) \(-0.06\) \(-0.02\) \(-0.01\) |
\(0.0396\%\) \(0.0027\%\) \(0.0005\%\) \(0.0002\%\) \(0.0001\%\) |
Error in Composite Simpson’s 1/3 Rule
The true error in a single application of Simpson’s 1/3rd Rule for \(\int_{a}^{b}{f(x)\text{dx}}\) is given by
\[E_{t}=-\frac{(b-a)^{5}}{2880} f^{(4)}(\zeta),\ a<\zeta<b\]
where \(\zeta\) is some point in \(\left\lbrack a,b \right\rbrack\) and \(f^{(4)}\) stands for the fourth derivative of the function.
What is the error then in the composite Simpson’s 1/3rd rule for \(\displaystyle \int_{a}^{b}{f(x){dx}}\)?
\[E_{t} = - \frac{(b - a)^{5}}{180n^{4}}{\overline{f}}^{\ (4)}\]
where
\[{\bar{f}}^{(4)} = \frac{\displaystyle \sum_{i = 1}^{\displaystyle \frac{n}{2}}{f^{(4)}(\zeta_{i})}}{\displaystyle \frac{n}{2}}\]
\[n =\text{number of segments}\]
Learning Objectives
After successful completion of this lesson, you should be able to:
1) derive the formula for the true error in the composite (also called multiple-segment) Simpson’s 1/3 rule of integration.
True Error in Composite Simpson’s 1/3 rule
The true error in a single application of Simpson’s 1/3rd Rule is given by
\[E_{t}=-\frac{(b-a)^{5}}{2880} f^{(4)}(\zeta), a<\zeta<b\]
The \(f^{(4)}\) in the true error expression stands for the fourth derivative of the function
In composite Simpson’s 1/3 rule, the error is the sum of the errors in each application of Simpson’s 1/3 rule. The error in the n segments Simpson’s 1/3rd Rule is given by
\[\begin{split} E_{1} &= - \frac{(x_{2} - x_{0})^{5}}{2880}f^{(4)}(\zeta_{1}),x_{0} < \zeta_{1} < x_{2}\\ &= - \frac{h^{5}}{90}f^{(4)}(\zeta_{1})\\ E_{2} &= - \frac{(x_{4} - x_{2})^{5}}{2880}f^{(4)}(\zeta_{2}),x_{2} < \zeta_{2} < x_{4}\\ &=-\frac{h^{5}}{90} f^{(4)}\left(\zeta_{2}\right)\\ &\ \ \ \ \ \ \ \ \ \vdots \\ E_{i} &=-\frac{\left(x_{2 i}-x_{2(i-1)}\right)^{5}}{2880} f^{(4)}\left(\zeta_{i}\right), \quad x_{2(i-1)}<\zeta_{i}<x_{2 i}\\ &=-\frac{h^{5}}{90} f^{(4)}\left(\zeta_{i}\right) \\ & \ \ \ \ \ \ \ \ \ \vdots \\ E_{\frac{n}{2}-1} &=-\frac{\left(x_{n-2}-x_{n-4}\right)^{5}}{2880} f^{(4)}\left(\zeta_{\frac{n}{2}-1}\right), x_{n-4}<\zeta_{\frac{n}{2}-1}<x_{n-2}\\ &=-\frac{h^{5}}{90} f^{(4)}\left(\zeta_{\frac{n}{2}-1}\right) \\ E_{\frac{n}{2}} &= - \frac{(x_{n} - x_{n - 2})^{5}}{2880}f^{(4)}\left( \zeta_{\frac{n}{2}} \right),x_{n - 2} < \zeta_{\frac{n}{2}} < x_{n} \end{split}\]
Hence, the total error in the composite Simpson’s 1/3 rule is
\[ =-\frac{h^{5}}{90} f^{(4)}\left(\zeta_{\frac{n}{2}}\right)\] \[ \begin{split} E_{t} &=\sum_{i=1}^{\frac{n}{2}} E_{i}\\ &=-\frac{h^{5}}{90} \sum_{i=1}^{\frac{n}{2}} f^{(4)}\left(\zeta_{i}\right)\\ &=-\frac{(b-a)^{5}}{90 n^{5}} \sum_{i=1}^{\frac{n}{2}} f^{(4)}\left(\zeta_{i}\right)\\ &= - \frac{(b - a)^{5}}{180n^{4}}\frac{\displaystyle \sum_{i = 1}^{\frac{n}{2}}{f^{(4)}(\zeta_{i})}}{\displaystyle \frac{n}{2}}, \end{split}\]
The term \(\frac{\displaystyle \sum_{i = 1}^{\displaystyle \frac{n}{2}}{f^{(4)}(\zeta_{i})}}{\displaystyle \frac{n}{2}}\ \) is an approximate average value of \(f^{(4)}(x),a < x < b\). Hence
\[E_{t} = - \frac{(b - a)^{5}}{180n^{4}}{\overline{f}}^{(4)}\]
where
\[{\bar{f}}^{(4)} = \frac{\displaystyle \sum_{i = 1}^{\displaystyle \frac{n}{2}}{f^{(4)}(\zeta_{i})}}{\displaystyle \frac{n}{2}}\]
Multiple Choice Test
(1). The highest order of polynomial integrand for which Simpson’s 1/3 rule of integration is exact is
(A) first
(B) second
(C) third
(D) fourth
(2). The value of \(\displaystyle \int_{0.2}^{2.2}{e^{x}\text{dx}}\) by using single-application (2-segments) Simpson’s \(1/3\) rule most nearly is
(A) \(7.8036\)
(B) \(7.8423\)
(C) \(8.4433\)
(D) \(10.246\)
(3). The value of \(\displaystyle \int_{0.2}^{2.2}{e^{x}\text{dx}}\) by using composite Simpson’s \(1/3\) rule with \(4\) segments most nearly is
(A) \(7.8036\)
(B) \(7.8062\)
(C) \(7.8423\)
(D) \(7.9655\)
(4). The velocity of a body is given by
\[\begin{split} v\left( t \right) &= 2t,\ 1 \leq t \leq 5\\ &= 5t^{2} + 3,\ 5 < t \leq 14 \end{split}\]
where \(t\) is given in seconds, and \(v\) is given in \(m/s\). Using single-application (two-segment) Simpson’s \(1/3\) rule, the distance in meters covered by the body from \(t = 2\) to \(t = 9\) seconds most nearly is
(A) \(949.33\)
(B) \(1039.7\)
(C) \(1200.5\)
(D) \(1442.0\)
(5). The value of \(\displaystyle \int_{3}^{19}{f\left( x \right)\text{dx}}\) by using single-application (2-segment) Simpson’s 1/3 rule is estimated as 702.039. The estimate of the same integral using composite Simpson’s 1/3 rule with \(4\) segments most nearly is
(A) \(\displaystyle 702.039 + \frac{8}{3}\left\lbrack 2f\left( 7 \right) - f\left( 11 \right) + 2f\left( 15 \right) \right\rbrack\)
(B) \(\displaystyle \frac{702.039}{2} + \frac{8}{3}\left\lbrack 2f\left( 7 \right) - f\left( 11 \right) + 2f\left( 15 \right) \right\rbrack\)
(C) \(\displaystyle 702.039 + \frac{8}{3}\left\lbrack 2f\left( 7 \right) + 2f\left( 15 \right) \right\rbrack\)
(D) \(\displaystyle \frac{702.039}{2} + \frac{8}{3}\left\lbrack 2f\left( 7 \right)2f\left( 15 \right) \right\rbrack\)
(6). The following data of the velocity of a body is given as a function of time.
| \(Time\ (s)\) | \(4\) | \(7\) | \(10\) | \(15\) |
|---|---|---|---|---|
| \(Velocity\ (m/s)\) | \(22\) | \(24\) | \(37\) | \(46\) |
The best estimate of the distance in meters covered by the body from \(t = 4\) to \(t = 15\) using combined Simpson’s \(1/3\) rule and the trapezoidal rule would be
(A) \(354.70\)
(B) \(362.50\)
(C) \(368.00\)
(D) \(378.80\)
For complete solution, go to
http://nm.mathforcollege.com/mcquizzes/07int/quiz_07int_simpson_solution.pdf
Problem Set
(1). Find the value of \(\displaystyle \int_{0.2}^{2.2}{xe^{x}dx}\) by using single-application composite Simpson’s 1/3 rule.
Answer: \(12.012\)
(2). Find the value of \(\displaystyle \int_{- 3.4}^{2.2}{x^{2}e^{x}dx}\) by using composite Simpson’s 1/3 rule. Use \(4\) segments.
Answer: \(24.418\)
(3). The upward velocity of a rocket is given by
\[v(t) = 200\ln(t + 1) - 10t,\ t > 0\]
where \(t\) is given in seconds and \(v\) is given in m/s.
a) Use single-application Simpson’s 1/3 rule to calculate the distance covered by the rocket from \(t = 0\) to \(t = 5\)s.
b) What is the true value of the distance covered by the rocket from \(t = 0\) to \(t = 5\)s?
c) What is the true error in part (a)?
d) What is the relative true error in part (a)?
e) What is the absolute relative true error in percentage for part (a).
f) Use composite Simpson’s 1/3rd rule to calculate the distance covered by the rocket from \(t = 0\) to \(t = 5\)s. Use \(4\) segments.
g) What is the absolute relative approximate error in percentage for part (e), assuming the previous approximation is from part (a)?
h) Based on the answer from part (g), how many significant digits are correct in the answer in part (g).
Answer: \(a)\ 1008.8\ \text{m} \ \ b)\ 1025.1\ \text{m} \ \ c)\ 16.309\ \text{m} \ \ d)\ 0.015910 \ \ e)\ 1.5910 \% \ \ f)\ 1022.8 \ \text{m} \ \ g)\ 1.3685\% \ \ h)\ 1\)
(4). The upward velocity of a body is given by
\[v(t) = \ln\left( \frac{160}{160 - 9t} \right),\ t > 0,\]
where \(t\) is given in seconds, and \(v\) in m/s. What is the distance covered by the body from \(t = 5\) to \(t = 9\) seconds? Use composite Simpson’s 1/3 rule with \(4\) segments.
Answer: \(2.0251\ \text{m}\)
(5). The true error for the single-application Simpson’s 1/3 rule used to calculate the approximate value of the integral \(\displaystyle \int_{a}^{b}{f(x)dx}\) is given by
\[E_{t} = - \frac{(b - a)^{5}}{2880}f^{(4)}(\xi),a \leq \xi \leq b\]
For the integral\(\displaystyle \int_{3}^{6}{4e^{2x}}dx\), what is the value of \(\xi\)?
Answer: \(\xi = 4.7088\)
(6). The true error for the single application Simpson’s 1/3 rule used to calculate the approximate value of the integral \(\displaystyle \int_{a}^{b}{f(x)dx}\) is given by
\[E_{t} = - \frac{(b - a)^{5}}{2880}f^{(4)}(\xi),a \leq \xi \leq b\]
For the integral\(\displaystyle \int_{3}^{6}{4e^{2x}}dx\),
a) find the lower and upper estimate of the true error based on the true error formula.
b) find the exact value of the integral.
c) verify that the estimates of part (a) bracket the true error.
Answer: \(a)\ E_{\text{upper}} = -878880,\ E_{\text{lower}} = -2178.5\ \ b)\ \text{Exact Integral} = 324700,\ \text{Approximate Integral} = 391140\)