Learning Lab

Example 1

Using Simpson’s rule, find the value of \[\begin{align*} \int_{1}^{3} & \left(e^{2x}\sin\left(x\right)\right)dx \end{align*}\] using \(2\) strips to an accuracy of \(3\) decimal places. This integral may be evaluated and is approximately equal to \(100.96243981\).

Solution:

Consider the figure below.

Simpson’s rule approximates the real function by a parabola (shown in green/black dotted lines).11 Note that three points are needed to fit a parabola. The area given by Simpson’s rule is the shaded area under the parabola. From the graph we can see that Simpson’s rule will overestimate the area under the real graph from \(x=1\) to \(x=2\) and underestimate the area from \(x=2\) to \(x=3.\)

We have \(n=2,\) \(a=1,\) \(b=3\) and22 The first strip is from \(x=1\) to \(x=2.\) The second is from \(x=2\) to \(x=3.\) \[\begin{align*} h & =\frac{b-a}{2}\\ & =\frac{3-1}{2}\\ & =1. \end{align*}\]

Calculating the three points and the value of the function at those points we have:33 Note that the number of points \(p\) and the number of strips \(n\) are related by \[\begin{align*} p & =n+1. \end{align*}\] We stop calculating points when \(x_{n}=b.\) In this example \(n=2\) and there are \(3=2+1\) points at which the function must be evaluated. They are \(x_{0},x_{1}\) and \(x_{2}.\)

\(x_{0}=a=1\) and \(y_{0}=e^{2\left(1\right)}\sin\left(1\right)=6.2176763.\)

\(x_{1}=a+h=1+1=2\) and \(y_{1}=e^{2\left(2\right)}\sin\left(2\right)=49.6459573.\)

\(x_{2}=a+2h=1+2=3\) and \(y_{3}=e^{2\left(3\right)}\sin\left(3\right)=56.9318746.\)

Now we substitute this in the Simpson’s rule formula \(\left(1\right).\) We get \[\begin{align*} \int_{1}^{3}\left(e^{2x}\sin\left(x\right)\right)dx & =\frac{1}{3}h\left[y_{0}+4y_{1}+y_{2}\right]\\ & =\frac{1}{3}\left(1\right)\left[6.2176763+4\left(49.6459573\right)+56.9318746\right]\\ & =\frac{1}{3}\left[6.2176763+4\left(49.6459573\right)+56.9318746\right]\\ & =87.2444601. \end{align*}\] The answer, to three decimal places, is \[\begin{align*} \int_{1}^{3}\left(e^{2x}\sin\left(x\right)\right)dx & =87.244. \end{align*}\] This is well under the value of \(100.96243981\). The percentage error is \[\begin{align*} \text{error} & =\frac{100.96243981-87.2444601}{100.96243981}\times100\\ & =13.6\%. \end{align*}\]

Example 2

Using Simpson’s rule, find the value of \[\begin{align*} \int_{1}^{3} & \left(e^{2x}\sin\left(x\right)\right)dx \end{align*}\] using \(4\) strips to an accuracy of \(3\) decimal places. This is exactly the same as Example 1 above except that we are using more strips.

Solution:

We have \(n=4,\) \(a=1,\) \(b=3\) and \[\begin{align*} h & =\frac{b-a}{4}\\ & =\frac{3-1}{4}\\ & =0.5\,. \end{align*}\]

Now calculate the points and the value of the function at those points.

\(x_{0}=a=1\) and \(y_{0}=e^{2\left(1\right)}\sin\left(1\right)=6.2176763.\)

\(x_{1}=a+h=1+0.5=1.5\) and \(y_{1}=e^{2\left(1.5\right)}\sin\left(1.5\right)=20.0352224\,.\)

\(x_{2}=a+2h=1+1=2\) and \(y_{3}=e^{2\left(2\right)}\sin\left(2\right)=49.6459573\,.\)

\(x_{3}=a+3h=1+1.5=2.5\) and \(y_{3}=e^{2\left(2.5\right)}\sin\left(2.5\right)=88.8211415\,.\)

\(x_{4}=a+4h=1+2=3\) and \(y_{4}=e^{2\left(3\right)}\sin\left(3\right)=56.9318746\,.\)

Now we substitute the \(y_{n}\) values in the Simpson’s rule formula \(\left(1\right).\) We get \[\begin{align*} \int_{1}^{3}\left(e^{2x}\sin\left(x\right)\right)dx & =\frac{1}{3}h\left[y_{0}+4\left(y_{1}+y_{3}\right)+2\left(y_{2}\right)+y_{4}\right]\\ & =\frac{0.5}{3}\left[6.2176763+4\left(20.0352224+88.8211415\right)\right.\\ & \quad\quad\left.+2\left(49.6459573\right)+56.9318746\right]\\ & =99.6444869\,. \end{align*}\] So the answer to three decimal places is \[\begin{align*} \int_{1}^{3}\left(e^{2x}\sin\left(x\right)\right)dx & =99.645\,. \end{align*}\] This is much closer to the actual value of \(100.96243981\). The percentage error is now \[\begin{align*} \text{error} & =\frac{100.96243981-99.6444869}{100.96243981}\times100\\ & =1.3\% \end{align*}\] which is a huge improvement over the result in Example 1.

Example 3

Using Simpson’s rule, find the value of \[\begin{align*} \int_{1}^{3} & \left(e^{2x}\sin\left(x\right)\right)dx \end{align*}\] using \(6\) strips to an accuracy of \(3\) decimal places. This is exactly the same as Example 1 and 2 above, except that we are using more strips.

Solution:

We have \(n=6,\ \) \(a=1,\) \(b=3\) and \[\begin{align*} h & =\frac{b-a}{6}\\ & =\frac{3-1}{6}\\ & =\frac{1}{3}\,. \end{align*}\]

Now calculate the values at the seven points:

\(x_{0}=a=1\) and \(y_{0}=e^{2\left(1\right)}\sin\left(1\right)=6.2176763.\)

\(x_{1}=a+h=1+1/3=4/3.\) So \(y_{1}=e^{2\left(4/3\right)}\sin\left(4/3\right)=13.9880487\,.\)

\(x_{2}=a+2h=1+2/3=5/3.\) So \(y_{3}=e^{2\left(5/3\right)}\sin\left(5/3\right)=27.9029025\,.\)

\(x_{3}=a+3h=1+1=2\,.\) So \(y_{3}=e^{2\left(2\right)}\sin\left(2\right)=49.6459573\,.\)

\(x_{4}=a+4h=1+4/3=7/3.\) So \(y_{4}=e^{2\left(7/3\right)}\sin\left(7/3\right)=76.8948872\,.\)

\(x_{5}=a+5h=1+5/3=8/3.\) So \(y_{5}=e^{2\left(8/3\right)}\sin\left(8/3\right)=94.7136211\,.\)

\(x_{6}=a+6h=1+2=3.\) So \(y_{6}=e^{2\left(3\right)}\sin\left(3\right)=56.9318746\,.\)

Now we substitute these values into the Simpson’s rule formula \(\left(1\right).\) We get \[\begin{align*} \int_{1}^{3}\left(e^{2x}\sin\left(x\right)\right)dx & =\frac{1}{3}h\left[y_{0}+4\left(y_{1}+y_{3}+y_{5}\right)+2\left(y_{2}+y_{4}\right)+y_{6}\right]\\ & =\frac{1}{3}\frac{1}{3}\left[6.2176763\right.\\ & \quad\quad+4\left(13.9880487+49.6459573+94.7136211\right)\\ & \quad\quad+2\left(27.9029025+76.8948872\right)+56.9318746\left.\!\right]\\ & =100.6817377\,. \end{align*}\] The answer is, to three decimal places, \[\begin{align*} \int_{1}^{3} & \left(e^{2x}\sin\left(x\right)\right)dx=100.682\,. \end{align*}\] The percentage error is now \[\begin{align*} \text{error} & =\frac{100.96243981-100.6817377}{100.96243981}\times100\\ & =0.3\%. \end{align*}\]

Answers

1. 45.00

2. 2.72