OBJECTIVES

When you finish this section, you should be able to:

1. Approximate an integral using the Trapezoidal Rule (p. 508)
2. Approximate an integral using Simpson’s Rule (p. 514)

RECALL

The area \(A\) of a trapezoid with width \(\Delta x\) and parallel heights \(l_{1}\) and \(l_{2}\) is \(A\) \(=\dfrac{1}{2}(l_{1}+l_{2}) \Delta x.\)

The Trapezoidal Rule

If a function \(f\) is continuous on the closed interval \([a,\,b ]\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int_{a}^{b}f(x)\,dx\approx \dfrac{b-a}{2n}\,\left[f(x_{0})+2f(x_{1})+2f(x_{2})+\cdots +2f(x_{n-1})+f(x_{n}) \right] } \]

where the closed interval \([a,b]\) is partitioned into \(n\) subintervals \( [x_{0}, x_{1}]\), \([x_{1}, x_{2}]\), \(\ldots ,\) \([x_{n-1}, x_{n}]\), each of width \(\Delta x=\dfrac{b-a}{n}\).

Proof

Assume \(f\) is nonnegative and continuous on \([a,b ].\) We seek the area \(A\) of the region bounded by the graph of \(f\), the \(x\)-axis, and the lines \(x=a\) and \(x=b.\) We partition the closed interval \([a,b]\) into \(n\) subintervals, \([a, x_{1}], [x_{1}, x_{2}], \ldots , [x_{i-1}, x_{i}], \ldots ,\) \( [x_{n-1}, b]\), each of width \(\Delta x=\dfrac{b-a}{n}\). The \(y\)-coordinates corresponding to the numbers \( x_{0}=a, x_{1}, x_{2}, \ldots , x_{i-1},\) \(x_{i}, \ldots , x_{n}=b\) are \(f(x_{0})\), \(f(x_{1})\), \(f(x_{2}), \ldots , f( x_{i-1}) ,\) \(f(x_{i}) , \ldots, f(x_{n-1})\) , \(f(x_{n})\). We connect each pair of consecutive points \((x_{i-1}, f( x_{i-1})) \) and \(\left( x_{i}, f(x_{i}) \right) \), \(i=1,2, \ldots , n,\) with a line segment to form \(n\) trapezoids. The trapezoid whose base is \(\Delta x=x_{i}-x_{i-1}=\dfrac{b-a}{n}\) has parallel sides of lengths \(f( x_{i-1}) \) and \(f(x_{i}) .\) Based on Figure 12, its area \(A_{i}\) is \[ A_{i}=\dfrac{1}{2}\left[f( x_{i-1}) +f(x_{i}) \right] \Delta x \]

The sum of the areas of the \(n\) trapezoids approximates the area \(A\) under the graph of \(f\). See Figure 13.

Figure 13 \(\sum\limits_{i=1}^{n} A_{i} \approx \int_{a}^{b} f( x) \,dx\)

Then \[ \begin{eqnarray*} A &\approx & \dfrac{1}{2}\,[f(x_{0})+f(x_{1})] \,\Delta x+ \dfrac{1}{2}\,[f(x_{1})+f(x_{2})] \,\Delta x+\cdots +\dfrac{ 1}{2}\,[f(x_{n-1}) +f(x_{n})] \,\Delta x \\[6pt] &\approx & \dfrac{1}{2}[f(x_{0})+2f(x_{1})+2f(x_{2})+\cdots +2f(x_{n-1})+f(x_{n})] \Delta x \end{eqnarray*} \]

Now substitute \(\Delta x=\dfrac{b-a}{n}\) to obtain the Trapezoidal Rule.

Approximating \(\displaystyle\int_{0}^{\pi }\sin x\, dx\) Using the Trapezoidal Rule

1. Use the Trapezoidal Rule with \(n=4\) and \(n=6\) to approximate \( \int_{0}^{\pi }\sin x\,dx\), rounded to three decimal places.
2. Compare each approximation to the exact value of \(\int_{0}^{\pi }\sin xdx\).

Solution  (a) We partition the interval \([0,\pi] \) into four subintervals, each of width \(\Delta x=\dfrac{\pi -0}{4}=\dfrac{\pi }{4}.\) \[ \left[ 0,\dfrac{\pi }{4}\right] \qquad \left[ \dfrac{\pi }{4},\dfrac{ \pi }{2}\right]\hbox{} \qquad \left[ \dfrac{\pi }{2},\dfrac{3\pi }{4}\right] \hbox{}\qquad \left[\dfrac{3\pi }{4},\pi \right] \]

The values of \(f( x) =\sin x\) corresponding to each endpoint are \[ f(0) =0 \!\qquad f\!\left( \dfrac{\pi }{4} \right) =\dfrac{\sqrt{2}}{2}\!\qquad f\!\left( \dfrac{\pi }{2} \right) =1\!\qquad f\!\left( \dfrac{3\pi }{4}\right) = \dfrac{\sqrt{2}}{2}\!\qquad f (\pi) =0 \]

See Figure 14. Now we use the Trapezoidal Rule: \[ \begin{eqnarray*} \int_{0}^{\pi }\sin x\,dx &\approx &\dfrac{\pi -0}{2\cdot 4}\,\left[ \sin (0)+2\,\sin \left( \dfrac{\pi }{4}\right) +2\,\sin \left( \dfrac{\pi }{2} \right) +2\,\sin \left( \dfrac{3\pi }{4}\right) +\sin (\pi )\right] \\[6pt] &=&\dfrac{\pi }{8}\,\left[ 0+2\left( \dfrac{\sqrt{2}}{2}\right) +2\left( 1\right) +2\left( \dfrac{\sqrt{2}}{2}\right) +0\right] =\dfrac{\pi }{8} \big( 2+2\sqrt{2}\big) \\[6pt] &=& \dfrac{\pi }{4}\big( 1+\sqrt{2}\big) \approx 1.896 \end{eqnarray*} \]

Figure 14 \(n=4\).

To approximate \(\int_{0}^{\pi }\sin x\,dx\) using six subintervals, we partition \([0,\pi] \) into six subintervals, each of width \( \Delta x=\dfrac{\pi -0}{6}=\dfrac{\pi }{6},\) namely, \[ \left[ 0,\dfrac{\pi }{6}\right]\! \qquad \left[ \dfrac{\pi }{6},\dfrac{ \pi }{3}\right]\!\qquad \left[ \dfrac{\pi }{3},\dfrac{\pi }{2}\right]\! \qquad \left[ \dfrac{\pi }{2},\dfrac{2\pi }{3}\right]\!\qquad \left[\dfrac{2\pi }{3},\dfrac{5\pi }{6}\right]\! \qquad \left[ \dfrac{ 5\pi }{6},\pi \right] \]

See Figure 15. Then we use the Trapezoidal Rule. \[ \begin{eqnarray*} \int_{0}^{\pi }\sin x\,dx &\approx & \dfrac{\pi }{2\cdot 6}\left[ 0+2\cdot \dfrac{1}{2}+2\cdot \dfrac{\sqrt{3}}{2}+2\cdot 1+2\cdot \dfrac{\sqrt{3}}{2} +2\cdot \dfrac{1}{2}+0\right] \\[5pt] &=&\dfrac{\pi }{12}(4+2\sqrt{3}) = \dfrac{\pi }{6}(2+\sqrt{3}) \approx 1.954 \end{eqnarray*} \]

Figure 15 \(n=6.\)

(b) The exact value of the integral is \[ \int_{0}^{\pi }\sin x \, dx = \big[-\cos x\big] _{0}^{\pi }=-\cos \pi +\cos 0=1+1=2 \]

The approximation using the Trapezoidal Rule with four subintervals underestimates the integral by \(0.104.\) The approximation using six subintervals underestimates the integral by \(0.046.\) Notice that the approximation using six subintervals is more accurate than the approximation using four subintervals.

Approximating \(\int_{1}^{2}\dfrac{e^{x}}{x}\, dx\) Using the Trapezoidal Rule

Use the Trapezoidal Rule with \(n=4\) and \(n=6\) to approximate \(\int_{1}^{2} \dfrac{e^{x}}{x}dx\). Express the answer rounded to three decimal places.

Solution  We begin by partitioning the interval \([1,2]\) into four subintervals, each of width \(\Delta x=\dfrac{2-1}{4}=\dfrac{1}{4}\): \[ \left[ 1,\,\dfrac{5}{4}\right]\qquad \left[ \dfrac{5}{4},\,\dfrac{3}{2} \right]\qquad \left[ \dfrac{3}{2},\,\dfrac{7}{4}\right]\qquad \left[ \dfrac{7}{4},\,2\right] \]

The values of \(f ( x) =\dfrac{e^{x}}{x}\) corresponding to each endpoint are \[ \begin{eqnarray*} f(1) &=& e \quad f\!\left( \dfrac{5}{4}\right) = \dfrac{4e^{5/4}}{5}\quad f\!\left( \dfrac{3}{2}\right) =\dfrac{2e^{3/2}}{3}\quad f\!\left( \dfrac{7}{4}\right) = \dfrac{4e^{7/4}}{7}\quad f( 2) =\dfrac{e^{2}}{2} \end{eqnarray*} \]

Then, using the Trapezoidal Rule, we get \[ \int_{1}^{2}\dfrac{e^{x}}{x}dx\approx \dfrac{1}{2\cdot 4}\,\left[ e+2\cdot \dfrac{4e^{5/4}}{5}+2\,\cdot \dfrac{2e^{3/2}}{3}+2\,\cdot \dfrac{4e^{7/4}}{7} +\dfrac{e^{2}}{2}\right] \approx 3.069 \]

To approximate \(\int_{1}^{2}\dfrac{e^{x}}{x}\,dx\) using six subintervals, we partition \([1,2]\) into six subintervals, each of width \(\Delta x= \dfrac{2-1}{6}=\dfrac{1}{6}\): \[ \left[ 1,\,\dfrac{7}{6}\right]\! \quad \left[ \dfrac{7}{6},\,\dfrac{4}{3} \right]\! \quad \left[ \dfrac{4}{3},\,\dfrac{3}{2}\right]\! \quad \left[ \dfrac{3}{2},\dfrac{5}{3}\right]\!\quad \left[ \dfrac{5}{3},\dfrac{11}{6} \right]\!\quad \left[ \dfrac{11}{6},2\right] \]

Then, using the Trapezoidal Rule, we get \[ \begin{eqnarray*} \int_{1}^{2}\dfrac{e^{x}}{x}dx\approx \dfrac{1}{2\cdot 6}\,\left[f(1)+2f\! \left( \dfrac{7}{6}\right) +2f\! \left( \dfrac{4}{3}\right) \right.&+& 2f\! \left( \dfrac{3}{2}\right) +2f\! \left( \dfrac{5}{3}\right) \\[5pt] &+& \left. 2f\!\left( \dfrac{11}{6}\right) +f(2) \right] \approx 3.063 \end{eqnarray*} \]

NOW WORK

Problem 9(a).

THEOREM Error in the Trapezoidal Rule

Let \(f\) be a function for which \(f^{\prime \prime}\) is continuous on some open interval containing \(a\) and \(b.\) The error in using the Trapezoidal Rule to approximate \(\int_{a}^{b}f( x) \,dx\) can be estimated* using the formula \[\bbox[5px, border:1px solid black, #F9F7ED] {{Error} \leq \dfrac{(b-a)^{3}M}{12n^{2}}} \]

where \(M\) is the absolute maximum value of \(\vert f^{\prime \prime} \vert\) on the closed interval \([a,b]\).

NOTE

The error formula gives an upper bound to the error. A derivation of the error formula, which uses an extension of the Mean Value Theorem, may be found in numerical analysis books.

Estimating the Error in Using the Trapezoidal Rule

Estimate the error that results from using the Trapezoidal Rule with \(n=4\) and with \(n=6\) to approximate \( \int_{1}^{2}\dfrac{e^{x}}{x}dx\). See Example 2.

Solution To estimate the error resulting from approximating \( \int_{1}^{2}\dfrac{e^{x}}{x}dx\) using the Trapezoidal Rule, we need to find the absolute maximum value of \(\vert f^{\prime \prime}\vert\) on the interval \([ 1,2] .\) We begin by finding \(f^{\prime \prime}\): \[ \begin{eqnarray*} f( x) &=&\dfrac{e^{x}}{x} \\[5pt] f^{\prime} (x) &=&\dfrac{d}{dx}\dfrac{e^{x}}{x}=\dfrac{xe^{x}-e^{x}}{x^{2}}= \dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}} \\[5pt] f^{\prime \prime} (x) &=&\dfrac{d}{dx}\left( \dfrac{e^{x}}{x}-\dfrac{e^{x}}{ x^{2}}\right) =\dfrac{d}{dx}\dfrac{e^{x}}{x}-\dfrac{d}{dx}\dfrac{e^{x}}{x^{2} }=\left( \dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}}\right) -\dfrac{ e^{x}x^{2}-2e^{x}x}{x^{4}}\\[5pt] &=&\dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}}-\dfrac{e^{x} }{x^{2}}+\dfrac{2e^{x}}{x^{3}} \\[5pt] &=&e^{x}\left( \dfrac{1}{x}-\dfrac{2}{x^{2}}+\dfrac{2}{x^{3}}\right) \end{eqnarray*} \]

Since \(\vert f^{\prime \prime}\vert\) is continuous on the interval \([1,2] ,\) the Extreme Value Theorem guarantees that \( \vert f^{\prime \prime}\vert\) has an absolute maximum on \([1,2]\). We find the absolute maximum of \(\vert f^{\prime \prime}\vert\) using graphing technology. As seen from Figure 17, the absolute maximum is at the left endpoint \(1.\) The absolute maximum value of \(f^{\prime \prime}\) is \(\left\vert e^{1}\left( \dfrac{1}{1}-\dfrac{ 2}{1^{2}}+\dfrac{2}{1^{3}}\right) \right\vert =e.\) So, \(M=e\).

Figure 17 The absolute maximum of \(\vert f^{\prime\prime} \vert\) on \([1,2 ]\) is at \(x=1.\)

When \(n=4,\) the error in using the Trapezoidal Rule is \[ \hbox{Error} \leq \dfrac{( b-a) ^{3}M}{12n^{2}}=\dfrac{(2-1)^{3}(e) }{(12)(4^{2})}=\dfrac{e}{192}\approx 0.014 \]

That is, \[ \begin{eqnarray*} 3.069-0.014 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.069+0.014 \\[5pt] 3.055 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.083 \end{eqnarray*} \]

When \(n=6,\) the error in using the Trapezoidal Rule is \[ \hbox{Error} \leq \dfrac{\left( b-a\right) ^{3}M}{12n^{2}}=\dfrac{(2-1)^{3}(e) }{(12)(6^{2})}=\dfrac{e}{432}\approx 0.006 \]

That is, \[ \begin{eqnarray*} 3.063-0.006 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.063+0.006 \nonumber \\[5pt] 3.057 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.069 \end{eqnarray*} \]

NOW WORK

Problem 9(b).

Obtaining a Desired Accuracy Using the Trapezoidal Rule

Find the number of subintervals \(n\) needed to guarantee that the Trapezoidal Rule approximates \(\int_{1}^{2}\dfrac{e^{x}}{x}dx\) correct to within 0.0001.

Solution To be sure that the approximation of \(\int_{1}^{2} \dfrac{e^{x}}{x}dx\) is correct to within 0.0001, we require the error be less than \(0.0001.\) That is, \[ \begin{eqnarray*} \hbox{Error} & \leq & \dfrac{(b-a)^{3}M}{12n^{2}}=\dfrac{(2-1)^{3}M}{12n^{2}}\underset{\underset{\color{#0066A7}{M=e}}{\color{#0066A7}{\uparrow}}} {=}\dfrac{e}{12n^{2}} \lt 0.0001 \\ n^{2} & \gt &\dfrac{e}{(0.0001) (12) }=\dfrac{e}{0.0012} \\ n & \gt &{\sqrt{\dfrac{e}{0.0012}}}\approx 47.6 \end{eqnarray*} \]

To ensure that the error is less than \(0.0001,\) we round up to \(48\) subintervals.

NOW WORK

Problem 9(c).

Using the Trapezoidal Rule with Empirical Data

A \(140\)-foot \(({\rm ft})\) tree trunk is cut into 20-ft logs. The diameter of each cross-sectional cut is measured and its area \(A\) recorded in the table below. (\(x\) is the distance in feet of the cut from the top of the trunk.)

Find the approximate volume of the tree trunk.

Solution The volume of the tree trunk is \(V=\int_{0}^{140}A ( x) \,dx\), where \(A( x)\) is the area of a slice at \(x.\) Since only eight data points are given, the function \(A( x)\) is not explicitly known. To approximate the volume, we partition the interval \([ 0,140]\) into seven subintervals, each of width \( \Delta x=\dfrac{140}{7}=20\). This partition corresponds to the given data. Using the Trapezoidal Rule, the approximate volume of the tree trunk is \[ \begin{eqnarray*} V= \int_{0}^{140}A( x) \,dx\approx \dfrac{1}{2}(20) \,[ A(0)+2\,A\,(20)&+&2\,A(40)+2\,A (60) +2\,A(80) \\[5pt] &+& 2\,A (100) +2\,A(120) +A\,(140)] \end{eqnarray*} \] \[ \begin{eqnarray*} V &\approx & 10\,[120+2(124)+2(128)+2(130)+2(132)+2(136)+2(144)+158]\\[5pt] &=& 18{,}660 \end{eqnarray*} \]

The volume of the tree trunk is approximately \(18{,}660{\rm ft^{3}}.\)

NOW WORK

Problem 25.

NOTE

With the Trapezoidal Rule, we approximate a function \(f\) with line segments (first degree polynomials), whereas with Simpson’s Rule, we approximate \(f\) with parabolic arcs (second degree polynomials).

ORIGINS

Simpson’s Rule is named after the British mathematician Thomas Simpson, who lived from 1710 to 1761.

Simpson’s Rule

If a function \(f\) is continuous on the closed interval \([a,\,b]\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{ \begin{array}{lll} \int_{a}^{b}f(x)\,dx \approx \dfrac{b-a}{3n} [ f(x_{0})&+&4 f(x_{1})+2 f(x_{2})+4 f(x_{3})+2f(x_{4}) \\[8pt] && + \cdots + 2f(x_{n-2})+4 f(x_{n-1})+f(x_{n})] \end{array} } \]

where the closed interval \([a,b]\) is partitioned into an even number \(n\) of subintervals \([x_{0}, x_{1}]\), \([x_{1}, x_{2}]\), \(\ldots ,\) \( [x_{n-1}, x_{n}]\), each of width \(\dfrac{b-a}{n}\).

Approximating an Integral Using Simpson’s Rule

Use Simpson’s Rule with \(n=4\) to approximate \(\int_{\pi }^{2\pi }\dfrac{\sin x}{x}dx\). Express the answer rounded to three decimal places.

Solution We partition the interval \([\pi ,2\pi ]\) into the four subintervals \[ \left[ \pi ,\dfrac{5\pi }{4}\right]\!\qquad \left[ \dfrac{5\pi }{4},\dfrac{ 3\pi }{2}\right]\!\qquad \left[ \dfrac{3\pi }{2},\dfrac{7\pi }{4}\right]\!\qquad \left[ \dfrac{7\pi }{4},2\pi \right] \]

each of width \(\dfrac{\pi }{4}.\) The value of \(f( x) =\dfrac{\sin x}{x}\) corresponding to each endpoint is \[ \begin{eqnarray*} f(\pi) &=& 0\ \ \ f\!\left( \dfrac{5\pi }{4}\right) = - \dfrac{2\sqrt{2}}{5\pi}\ \ \ f\!\left( \dfrac{3\pi }{2} \right) =-\dfrac{2}{3\pi}\ \ \ f\!\left( \dfrac{7\pi }{4}\right) = -\dfrac{2\sqrt{2}}{7\pi}\quad f(2\pi) =0 \end{eqnarray*} \]

Then using Simpson’s Rule, we get \[ \begin{eqnarray*} \int_{\pi }^{2\pi }\dfrac{\sin x}{x}\,dx &\approx &\dfrac{2\pi -\pi }{3\cdot 4} \left[ f(\pi )+4\, f\! \left( \dfrac{5\pi }{4}\right) +2 f\! \left( \dfrac{3\pi }{2}\right) +4 f\! \left( \dfrac{7\pi }{4}\right) +f( 2\pi ) \right] \nonumber \\[7pt] &= &\left( \dfrac{\pi }{12}\right) \left[ 0+4\left( -\dfrac{2\sqrt{2}}{ 5\pi }\right) +2\left( -\dfrac{2}{3\pi }\right) +4\left( -\dfrac{2\sqrt{2}}{ 7\pi }\right) +0\right] \approx -0.434 \end{eqnarray*} \]

THEOREM Error in Simpson’s Rule

Let \(f\) be a function for which \(f^{(4)}\) is continuous on an open interval containing \(a\) and \(b.\) The error in using Simpson’s Rule to approximate \(\int_{a}^{b}f(x)\,dx\) can be estimated by the formula \[\bbox[5px, border:1px solid black, #F9F7ED]{ \hbox{Error} \leq \frac{(b-a)^{5}M}{180n^{4}}} \]

where \(M\) is the absolute maximum value of \(\vert f^{\,(4)}(x)\vert\) on the closed interval \([a,b]\).

NOTE

A CAS can be used to find higher-order derivatives.

Estimating the Error in Using Simpson's Rule

Estimate the error that results from using Simpson's Rule with \(n =4\) to approximate \( \int_{\pi }^{2\pi }\frac{\sin x}{x}dx\). (See Example 6.)

Solution To estimate the error, we need to find the absolute maximum value of \(\vert f^{(4) }\vert\) on the interval \([\pi ,2\pi ]\). We begin by finding \(f^{(4) }\): \[ \begin{eqnarray*} f(x) &=&\dfrac{\sin x}{x} \\[5pt] f^{\prime} (x) &=&\frac{x\cos x-\sin x}{x^{2}}=\frac{\cos x}{x}-\frac{\sin x}{x^{2}} \\ && \\ f^{\prime \prime} (x) &=&\left[ \frac{-x\sin x-\cos x}{x^{2}} \right] -\left[ \frac{x^{2}\cos x-2x\sin x}{x^{4}}\right]\\[6pt] & = & \frac{2 \sin x}{x^{3}} -\frac{\sin x}{x}-\frac{2\cos x}{x^{2}} \\[15pt] f^{\prime \prime \prime} (x) &=&\left[ \frac{2x^{3}\cos x-6x^{2}\sin x}{x^{6}}\right] -\left[ \frac{x\cos x-\sin x}{x^{2}}\right] - \left[ \frac{-2x^{2}\sin x-4x\cos x}{x^{4}}\right] \\[6pt] &=&\frac{2\cos x}{x^{3}}-\frac{6\sin x}{x^{4}}-\frac{\cos x}{x}+\frac{ \sin x}{x^{2}}+\frac{2\sin x}{x^{2}}+\frac{4\cos x}{x^{3}}\\[6pt] &=&\frac{6 \cos x}{x^{3}}-\frac{\cos x}{x}+\frac{3 \sin x}{x^{4}} \\[6pt] && \\ f^{\,(4) }(x) &=&\left[ \frac{-6x^{3}\sin x-18x^{2}\cos x}{ x^{6}}\right] -\left[ \frac{-x\sin x-\cos x}{x^{2}}\right] \\[8pt] &&+\,\left[ \frac{3x^{2}\cos x-6x\sin x}{x^{4}}\right] -\left[ \frac{6x^{4}\cos x-24x^{3}\sin x}{x^{8}}\right] \\[8pt] &=&\frac{4 \cos x}{x^{2}}-\frac{24 \cos x}{x^{4}}+\frac{\sin x}{x}-\frac{ 12 \sin x}{x^{3}}+\frac{24 \sin x}{x^{5}} \end{eqnarray*} \]

We use graphing technology to find the absolute maximum value of \(\vert f^{(4)}\vert\). See Figure 20.

Figure 20 \(y=\vert f^{(4)}\vert\) on the interval \([\pi, 2\pi]\).

Since on the interval \([\pi ,2\pi]\) the maximum value of \(\vert f^{(4)}\vert \lt 0.176,\) we use \(M=0.176\). Then the error that results from using Simpson's Rule to approximate \(\int_{\pi }^{2\pi }\frac{\sin x}{x}dx\) is \[ \hbox{Error} \leq \frac{M(b-a)^{5}}{180n^{4}}=\frac{0.176(2\pi -\pi )^{5}}{180\cdot 4^{4}}\approx 0.001 \]

That is, \[ \begin{eqnarray*} -0.434-0.001 & \leq& \int_{\pi }^{2\pi }\frac{\sin x}{x}dx \leq -0.434+0.001 \\[5pt] -0.435 & \leq& \int_{\pi }^{2\pi }\frac{\sin x}{x}dx \leq -0.433 \end{eqnarray*} \]

NOW WORK

Problem 13(a) and (b).

Obtaining a Desired Accuracy Using Simpson’s Rule

Find the number of subintervals needed to guarantee that Simpson’s Rule approximates \(\int_{\pi }^{2\pi }\frac{\sin x}{x}\, dx\) correct to within 0.0001.

Solution To be sure that the approximation of \(\int_{\pi }^{2\pi } \frac{\sin x}{x}dx\) is correct to within 0.0001, we require the error be less than \(0.0001.\) That is, \[ \begin{eqnarray*} \hbox{Error} & \leq &\dfrac{(b-a)^{5}M}{180n^{4}}=\dfrac{(2\pi -\pi )^{5}M}{180n^{4}} &&\underset{\underset{\color{#0066A7}{M=0.176}}{\color{#0066A7}{{\uparrow }}}}{=} \dfrac{( \pi ^{5}) (0.176) }{180n^{4}} \lt 0.0001 \\[-8pt] n^{4} &>&\dfrac{0.176\pi ^{5}}{(0.0001) \left( 180\right) } \\[6pt] n &>&\sqrt[4]{2992.192}\approx 7.396 \end{eqnarray*} \]

Since Simpson's Rule requires \(n\) to be even, eight subintervals are needed to guarantee that Simpson's Rule approximates \(\int_{\pi }^{2\pi }\frac{\sin x }{x} dx\) correct to within 0.0001.

NOW WORK

Problem 13(c).