{"id":1599,"date":"2021-07-22T16:41:29","date_gmt":"2021-07-22T16:41:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1599"},"modified":"2022-03-19T04:18:19","modified_gmt":"2022-03-19T04:18:19","slug":"simpsons-rule","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/simpsons-rule\/","title":{"raw":"Simpson's Rule","rendered":"Simpson&#8217;s Rule"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use Simpson\u2019s rule to approximate the value of a definite integral to a given accuracy<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165041836263\">With the midpoint rule, we estimated areas of regions under curves by using rectangles. In a sense, we approximated the curve with piecewise constant functions. With the trapezoidal rule, we approximated the curve by using piecewise linear functions. What if we were, instead, to approximate a curve using piecewise quadratic functions? With <span data-type=\"term\">Simpson\u2019s rule<\/span>, we do just this. We partition the interval into an even number of subintervals, each of equal width. Over the first pair of subintervals we approximate [latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{2}}f\\left(x\\right)dx[\/latex] with [latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{2}}p\\left(x\\right)dx[\/latex], where [latex]p\\left(x\\right)=A{x}^{2}+Bx+C[\/latex] is the quadratic function passing through [latex]\\left({x}_{0},f\\left({x}_{0}\\right)\\right)[\/latex], [latex]\\left({x}_{1},f\\left({x}_{1}\\right)\\right)[\/latex], and [latex]\\left({x}_{2},f\\left({x}_{2}\\right)\\right)[\/latex] (Figure 4). Over the next pair of subintervals we approximate [latex]{\\displaystyle\\int }_{{x}_{2}}^{{x}_{4}}f\\left(x\\right)dx[\/latex] with the integral of another quadratic function passing through [latex]\\left({x}_{2},f\\left({x}_{2}\\right)\\right)[\/latex], [latex]\\left({x}_{3},f\\left({x}_{3}\\right)\\right)[\/latex], and [latex]\\left({x}_{4},f\\left({x}_{4}\\right)\\right)[\/latex]. This process is continued with each successive pair of subintervals.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_06_004\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233841\/CNX_Calc_Figure_07_06_004.jpg\" alt=\"This figure has two graphs, both of the same non-negative function in the first quadrant. The function increases and decreases. The quadrant is divided into a grid. The first graph, beginning on the x-axis at the point labeled x sub 0, there are trapezoids shaded whose heights are represented by the function p(x), which is a curve following an approximate path of the original graph. The x-axis is scaled by increments of x sub 0, x sub 1, x sub 2. The second graph has on the x-axis at the point labeled x sub 0. There are shaded regions under the curve, divided by x sub 0, x sub 1, x sub 2, x sub 3, and x sub 4. The curve is sectioned into two different parts above the shaded areas. These two parts are labeled p sub 1(x) and p sub 2(x).\" width=\"975\" height=\"286\" data-media-type=\"image\/jpeg\" \/> Figure 4. With Simpson\u2019s rule, we approximate a definite integral by integrating a piecewise quadratic function.[\/caption]<\/figure>\r\n<p id=\"fs-id1165040644477\">To understand the formula that we obtain for Simpson\u2019s rule, we begin by deriving a formula for this approximation over the first two subintervals. As we go through the derivation, we need to keep in mind the following relationships:<\/p>\r\n\r\n<div id=\"fs-id1165040644485\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left({x}_{0}\\right)=p\\left({x}_{0}\\right)=A{x}_{0}{}^{2}+B{x}_{0}+C\\hfill \\\\ f\\left({x}_{1}\\right)=p\\left({x}_{1}\\right)=A{x}_{1}{}^{2}+B{x}_{1}+C\\hfill \\\\ f\\left({x}_{2}\\right)=p\\left({x}_{2}\\right)=A{x}_{2}{}^{2}+B{x}_{2}+C\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040777282\">[latex]{x}_{2}-{x}_{0}=2\\Delta x[\/latex], where [latex]\\Delta x[\/latex] is the length of a subinterval.<\/p>\r\n\r\n<div id=\"fs-id1165042288603\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{x}_{2}+{x}_{0}=2{x}_{1},\\:\\text{since}\\:{x}_{1}=\\frac{\\left({x}_{2}+{x}_{0}\\right)}{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041782442\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165041782445\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{{x}_{0}}^{{x}_{2}}f(x)dx}&amp; \\approx {\\displaystyle\\int _{{x}_{0}}^{{x}_{2}}p(x)dx}\\hfill &amp; &amp; &amp; \\\\ &amp; ={\\displaystyle\\int _{{x}_{0}}^{{x}_{2}}(A{x}^{2}+Bx+C)dx}\\hfill &amp; &amp; &amp; \\\\ &amp; =\\frac{A}{3}{x}^{3}+\\frac{B}{2}{x}^{2}+Cx|\\begin{array}{c}{}^{{x}_{2}} \\\\ {}_{{x}_{0}}\\end{array}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\frac{A}{3}({x}_{2}{}^{3}-{x}_{0}{}^{3})+\\frac{B}{2}({x}_{2}{}^{2}-{x}_{0}{}^{2})+C({x}_{2}-{x}_{0})\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\frac{A}{3}({x}_{2}-{x}_{0})({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2})\\hfill &amp; &amp; &amp;\r\n\\\\ &amp; +\\frac{B}{2}({x}_{2}-{x}_{0})({x}_{2}+{x}_{0})+C({x}_{2}-{x}_{0})\\hfill &amp; &amp; &amp; \\\\ &amp; =\\frac{{x}_{2}-{x}_{0}}{6}(2A({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2})+3B({x}_{2}+{x}_{0})+6C)\\hfill &amp; &amp; &amp; \\text{Factor out}\\frac{{x}_{2}-{x}_{0}}{6}.\\hfill \\\\ &amp; =\\frac{\\Delta x}{3}((Ax_{2}^{2}+B{x}_{2}+C)+(A{x}_{0}{}^{2}+B{x}_{0}+C)\\hfill &amp; &amp; &amp; \\\\ &amp; +A({x}_{2}{}^{2}+2{x}_{2}{x}_{0}+{x}_{0}{}^{2})+2B({x}_{2}+{x}_{0})+4C)\\hfill &amp; &amp; &amp; \\\\ &amp; =\\frac{\\Delta x}{3}(f({x}_{2})+f({x}_{0})+A{({x}_{2}+{x}_{0})}^{2}+2B({x}_{2}+{x}_{0})+4C)\\hfill &amp; &amp; &amp; \\text{Rearrange the terms.}\\hfill\r\n\\\\ &amp; &amp; &amp; &amp; \\begin{array}{c}\\text{Factor and substitute.}\\hfill \\\\ f({x}_{2})=A{x}_{0}{}^{2}+B{x}_{0}+C\\text{and}\\hfill \\\\ f({x}_{0})=A{x}_{0}{}^{2}+B{x}_{0}+C.\\hfill \\end{array}\\hfill \\\\ &amp; =\\frac{\\Delta x}{3}(f({x}_{2})+f({x}_{0})+A{(2{x}_{1})}^{2}+2B(2{x}_{1})+4C)\\hfill &amp; &amp; &amp; \\text{Substitute}{x}_{2}+{x}_{0}=2{x}_{1}.\\hfill \\\\ &amp; =\\frac{\\Delta x}{3}(f({x}_{2})+4f({x}_{1})+f({x}_{0})).\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Expand and substitute}\\hfill \\\\ f({x}_{1})=A{x}_{1}{}^{2}+B{x}_{1}\\text{+}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040770954\">If we approximate [latex]{\\displaystyle\\int }_{{x}_{2}}^{{x}_{4}}f\\left(x\\right)dx[\/latex] using the same method, we see that we have<\/p>\r\n\r\n<div id=\"fs-id1165041832972\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{4}}f\\left(x\\right)dx\\approx \\frac{\\Delta x}{3}\\left(f\\left({x}_{4}\\right)+4f\\left({x}_{3}\\right)+f\\left({x}_{2}\\right)\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042311586\">Combining these two approximations, we get<\/p>\r\n\r\n<div id=\"fs-id1165042092359\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{4}}f\\left(x\\right)dx=\\frac{\\Delta x}{3}\\left(f\\left({x}_{0}\\right)+4f\\left({x}_{1}\\right)+2f\\left({x}_{2}\\right)+4f\\left({x}_{3}\\right)+f\\left({x}_{4}\\right)\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041921702\">The pattern continues as we add pairs of subintervals to our approximation. The general rule may be stated as follows.<\/p>\r\n\r\n<div id=\"fs-id1165041921707\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Simpson\u2019s Rule<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165040721553\">Assume that [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right][\/latex]. Let <em data-effect=\"italics\">n<\/em> be a positive even integer and [latex]\\Delta x=\\frac{b-a}{n}[\/latex]. Let [latex]\\left[a,b\\right][\/latex] be divided into [latex]n[\/latex] subintervals, each of length [latex]\\Delta x[\/latex], with endpoints at [latex]P=\\left\\{{x}_{0},{x}_{1},{x}_{2},\\ldots ,{x}_{n}\\right\\}[\/latex]. Set<\/p>\r\n\r\n<div id=\"fs-id1165041953720\" style=\"text-align: center;\" data-type=\"equation\">[latex]{S}_{n}=\\frac{\\Delta x}{3}\\left(f\\left({x}_{0}\\right)+4f\\left({x}_{1}\\right)+2f\\left({x}_{2}\\right)+4f\\left({x}_{3}\\right)+2f\\left({x}_{4}\\right)+\\cdots +2f\\left({x}_{n - 2}\\right)+4f\\left({x}_{n - 1}\\right)+f\\left({x}_{n}\\right)\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042243662\">Then,<\/p>\r\n\r\n<div id=\"fs-id1165042243666\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\text{+}\\infty }{\\text{lim}}{S}_{n}={\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165042108863\">Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson\u2019s rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that [latex]{S}_{2n}=\\left(\\frac{2}{3}\\right){M}_{n}+\\left(\\frac{1}{3}\\right){T}_{n}[\/latex].<\/p>\r\n<p id=\"fs-id1165042264461\">It is also possible to put a bound on the error when using Simpson\u2019s rule to approximate a definite integral. The bound in the error is given by the following rule:<\/p>\r\n\r\n<div id=\"fs-id1165042264468\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Rule: Error Bound for Simpson\u2019s Rule<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165041836425\">Let [latex]f\\left(x\\right)[\/latex] be a continuous function over [latex]\\left[a,b\\right][\/latex] having a fourth derivative, [latex]{f}^{\\left(4\\right)}\\left(x\\right)[\/latex], over this interval. If [latex]M[\/latex] is the maximum value of [latex]|{f}^{\\left(4\\right)}\\left(x\\right)|[\/latex] over [latex]\\left[a,b\\right][\/latex], then the upper bound for the error in using [latex]{S}_{n}[\/latex] to estimate [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1165042109950\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\text{Error in}{S}_{n}\\le \\frac{M{\\left(b-a\\right)}^{5}}{180{n}^{4}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042217547\" data-type=\"example\">\r\n<div id=\"fs-id1165042217549\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042217551\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Applying Simpson\u2019s Rule 1<\/h3>\r\n<div id=\"fs-id1165042217551\" data-type=\"problem\">\r\n<p id=\"fs-id1165042217557\">Use [latex]{S}_{2}[\/latex] to approximate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{3}dx[\/latex]. Estimate a bound for the error in [latex]{S}_{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1165041848372\" data-type=\"solution\">\r\n<p id=\"fs-id1165041848374\">Since [latex]\\left[0,1\\right][\/latex] is divided into two intervals, each subinterval has length [latex]\\Delta x=\\frac{1 - 0}{2}=\\frac{1}{2}[\/latex]. The endpoints of these subintervals are [latex]\\left\\{0,\\frac{1}{2},1\\right\\}[\/latex]. If we set [latex]f\\left(x\\right)={x}^{3}[\/latex], then<\/p>\r\n<p id=\"fs-id1165041962775\">[latex]{S}_{4}=\\frac{1}{3}\\cdot \\frac{1}{2}\\left(f\\left(0\\right)+4f\\left(\\frac{1}{2}\\right)+f\\left(1\\right)\\right)=\\frac{1}{6}\\left(0+4\\cdot \\frac{1}{8}+1\\right)=\\frac{1}{4}[\/latex]. Since [latex]{f}^{\\left(4\\right)}\\left(x\\right)=0[\/latex] and consequently [latex]M=0[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1167793627790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Error in }{S}_{2}\\le \\frac{0{\\left(1\\right)}^{5}}{180\\cdot {2}^{4}}=0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041971466\">This bound indicates that the value obtained through Simpson\u2019s rule is exact. A quick check will verify that, in fact, [latex]{\\displaystyle\\int }_{0}^{1}{x}^{3}dx=\\frac{1}{4}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042199249\" data-type=\"example\">\r\n<div id=\"fs-id1165042199251\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042199253\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Applying Simpson\u2019s Rule 2<\/h3>\r\n<div id=\"fs-id1165042199253\" data-type=\"problem\">\r\n<p id=\"fs-id1165042199259\">Use [latex]{S}_{6}[\/latex] to estimate the length of the curve [latex]y=\\frac{1}{2}{x}^{2}[\/latex] over [latex]\\left[1,4\\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n<div id=\"fs-id1165040668581\" data-type=\"solution\">\r\n<p id=\"fs-id1165040668583\">The length of [latex]y=\\frac{1}{2}{x}^{2}[\/latex] over [latex]\\left[1,4\\right][\/latex] is [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{1+{x}^{2}}dx[\/latex]. If we divide [latex]\\left[1,4\\right][\/latex] into six subintervals, then each subinterval has length [latex]\\Delta x=\\frac{4 - 1}{6}=\\frac{1}{2}[\/latex], and the endpoints of the subintervals are [latex]\\left\\{1,\\frac{3}{2},2,\\frac{5}{2},3,\\frac{7}{2},4\\right\\}[\/latex]. Setting [latex]f\\left(x\\right)=\\sqrt{1+{x}^{2}}[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165040758286\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{6}=\\frac{1}{3}\\cdot \\frac{1}{2}\\left(f\\left(1\\right)+4f\\left(\\frac{3}{2}\\right)+2f\\left(2\\right)+4f\\left(\\frac{5}{2}\\right)+2f\\left(3\\right)+4f\\left(\\frac{7}{2}\\right)+f\\left(4\\right)\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042296653\">After substituting, we have<\/p>\r\n\r\n<div id=\"fs-id1165042296656\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {S}_{6}&amp; =\\frac{1}{6}\\left(1.4142+4\\cdot 1.80278+2\\cdot 2.23607+4\\cdot 2.69258+2\\cdot 3.16228+4\\cdot 3.64005+4.12311\\right)\\hfill \\\\ &amp; \\approx 8.14594.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example:\u00a0Applying Simpson\u2019s Rule 2[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XZ_mr5QqQNA?controls=0&amp;start=290&amp;end=543&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.6.5_290to543_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.6.5\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1165040796317\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165040796321\" data-type=\"exercise\">\r\n<div id=\"fs-id1165041899066\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165041899066\" data-type=\"problem\">\r\n<p id=\"fs-id1165041899068\">Use [latex]{S}_{2}[\/latex] to estimate [latex]{\\displaystyle\\int }_{1}^{2}\\frac{1}{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558499\"]Hint[\/reveal-answer]\r\n\r\n[hidden-answer a=\"44558499\"]\r\n<div id=\"fs-id1165040638592\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165042002716\">[latex]{S}_{2}=\\left(\\frac{1}{3}\\Delta x\\left(f\\left({x}_{0}\\right)+4f\\left({x}_{1}\\right)+f\\left({x}_{2}\\right)\\right)\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558599\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"44558599\"]\r\n<div id=\"fs-id1165040638577\" data-type=\"solution\">\r\n<p id=\"fs-id1165040638579\">[latex]\\frac{25}{36}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1165042276347\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5578[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use Simpson\u2019s rule to approximate the value of a definite integral to a given accuracy<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165041836263\">With the midpoint rule, we estimated areas of regions under curves by using rectangles. In a sense, we approximated the curve with piecewise constant functions. With the trapezoidal rule, we approximated the curve by using piecewise linear functions. What if we were, instead, to approximate a curve using piecewise quadratic functions? With <span data-type=\"term\">Simpson\u2019s rule<\/span>, we do just this. We partition the interval into an even number of subintervals, each of equal width. Over the first pair of subintervals we approximate [latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{2}}f\\left(x\\right)dx[\/latex] with [latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{2}}p\\left(x\\right)dx[\/latex], where [latex]p\\left(x\\right)=A{x}^{2}+Bx+C[\/latex] is the quadratic function passing through [latex]\\left({x}_{0},f\\left({x}_{0}\\right)\\right)[\/latex], [latex]\\left({x}_{1},f\\left({x}_{1}\\right)\\right)[\/latex], and [latex]\\left({x}_{2},f\\left({x}_{2}\\right)\\right)[\/latex] (Figure 4). Over the next pair of subintervals we approximate [latex]{\\displaystyle\\int }_{{x}_{2}}^{{x}_{4}}f\\left(x\\right)dx[\/latex] with the integral of another quadratic function passing through [latex]\\left({x}_{2},f\\left({x}_{2}\\right)\\right)[\/latex], [latex]\\left({x}_{3},f\\left({x}_{3}\\right)\\right)[\/latex], and [latex]\\left({x}_{4},f\\left({x}_{4}\\right)\\right)[\/latex]. This process is continued with each successive pair of subintervals.<\/p>\n<figure id=\"CNX_Calc_Figure_07_06_004\"><figcaption><\/figcaption><div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233841\/CNX_Calc_Figure_07_06_004.jpg\" alt=\"This figure has two graphs, both of the same non-negative function in the first quadrant. The function increases and decreases. The quadrant is divided into a grid. The first graph, beginning on the x-axis at the point labeled x sub 0, there are trapezoids shaded whose heights are represented by the function p(x), which is a curve following an approximate path of the original graph. The x-axis is scaled by increments of x sub 0, x sub 1, x sub 2. The second graph has on the x-axis at the point labeled x sub 0. There are shaded regions under the curve, divided by x sub 0, x sub 1, x sub 2, x sub 3, and x sub 4. The curve is sectioned into two different parts above the shaded areas. These two parts are labeled p sub 1(x) and p sub 2(x).\" width=\"975\" height=\"286\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. With Simpson\u2019s rule, we approximate a definite integral by integrating a piecewise quadratic function.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165040644477\">To understand the formula that we obtain for Simpson\u2019s rule, we begin by deriving a formula for this approximation over the first two subintervals. As we go through the derivation, we need to keep in mind the following relationships:<\/p>\n<div id=\"fs-id1165040644485\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left({x}_{0}\\right)=p\\left({x}_{0}\\right)=A{x}_{0}{}^{2}+B{x}_{0}+C\\hfill \\\\ f\\left({x}_{1}\\right)=p\\left({x}_{1}\\right)=A{x}_{1}{}^{2}+B{x}_{1}+C\\hfill \\\\ f\\left({x}_{2}\\right)=p\\left({x}_{2}\\right)=A{x}_{2}{}^{2}+B{x}_{2}+C\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040777282\">[latex]{x}_{2}-{x}_{0}=2\\Delta x[\/latex], where [latex]\\Delta x[\/latex] is the length of a subinterval.<\/p>\n<div id=\"fs-id1165042288603\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{x}_{2}+{x}_{0}=2{x}_{1},\\:\\text{since}\\:{x}_{1}=\\frac{\\left({x}_{2}+{x}_{0}\\right)}{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041782442\">Thus,<\/p>\n<div id=\"fs-id1165041782445\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{{x}_{0}}^{{x}_{2}}f(x)dx}& \\approx {\\displaystyle\\int _{{x}_{0}}^{{x}_{2}}p(x)dx}\\hfill & & & \\\\ & ={\\displaystyle\\int _{{x}_{0}}^{{x}_{2}}(A{x}^{2}+Bx+C)dx}\\hfill & & & \\\\ & =\\frac{A}{3}{x}^{3}+\\frac{B}{2}{x}^{2}+Cx|\\begin{array}{c}{}^{{x}_{2}} \\\\ {}_{{x}_{0}}\\end{array}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\frac{A}{3}({x}_{2}{}^{3}-{x}_{0}{}^{3})+\\frac{B}{2}({x}_{2}{}^{2}-{x}_{0}{}^{2})+C({x}_{2}-{x}_{0})\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\frac{A}{3}({x}_{2}-{x}_{0})({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2})\\hfill & & &  \\\\ & +\\frac{B}{2}({x}_{2}-{x}_{0})({x}_{2}+{x}_{0})+C({x}_{2}-{x}_{0})\\hfill & & & \\\\ & =\\frac{{x}_{2}-{x}_{0}}{6}(2A({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2})+3B({x}_{2}+{x}_{0})+6C)\\hfill & & & \\text{Factor out}\\frac{{x}_{2}-{x}_{0}}{6}.\\hfill \\\\ & =\\frac{\\Delta x}{3}((Ax_{2}^{2}+B{x}_{2}+C)+(A{x}_{0}{}^{2}+B{x}_{0}+C)\\hfill & & & \\\\ & +A({x}_{2}{}^{2}+2{x}_{2}{x}_{0}+{x}_{0}{}^{2})+2B({x}_{2}+{x}_{0})+4C)\\hfill & & & \\\\ & =\\frac{\\Delta x}{3}(f({x}_{2})+f({x}_{0})+A{({x}_{2}+{x}_{0})}^{2}+2B({x}_{2}+{x}_{0})+4C)\\hfill & & & \\text{Rearrange the terms.}\\hfill  \\\\ & & & & \\begin{array}{c}\\text{Factor and substitute.}\\hfill \\\\ f({x}_{2})=A{x}_{0}{}^{2}+B{x}_{0}+C\\text{and}\\hfill \\\\ f({x}_{0})=A{x}_{0}{}^{2}+B{x}_{0}+C.\\hfill \\end{array}\\hfill \\\\ & =\\frac{\\Delta x}{3}(f({x}_{2})+f({x}_{0})+A{(2{x}_{1})}^{2}+2B(2{x}_{1})+4C)\\hfill & & & \\text{Substitute}{x}_{2}+{x}_{0}=2{x}_{1}.\\hfill \\\\ & =\\frac{\\Delta x}{3}(f({x}_{2})+4f({x}_{1})+f({x}_{0})).\\hfill & & & \\begin{array}{c}\\text{Expand and substitute}\\hfill \\\\ f({x}_{1})=A{x}_{1}{}^{2}+B{x}_{1}\\text{+}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040770954\">If we approximate [latex]{\\displaystyle\\int }_{{x}_{2}}^{{x}_{4}}f\\left(x\\right)dx[\/latex] using the same method, we see that we have<\/p>\n<div id=\"fs-id1165041832972\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{4}}f\\left(x\\right)dx\\approx \\frac{\\Delta x}{3}\\left(f\\left({x}_{4}\\right)+4f\\left({x}_{3}\\right)+f\\left({x}_{2}\\right)\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042311586\">Combining these two approximations, we get<\/p>\n<div id=\"fs-id1165042092359\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{{x}_{0}}^{{x}_{4}}f\\left(x\\right)dx=\\frac{\\Delta x}{3}\\left(f\\left({x}_{0}\\right)+4f\\left({x}_{1}\\right)+2f\\left({x}_{2}\\right)+4f\\left({x}_{3}\\right)+f\\left({x}_{4}\\right)\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041921702\">The pattern continues as we add pairs of subintervals to our approximation. The general rule may be stated as follows.<\/p>\n<div id=\"fs-id1165041921707\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Simpson\u2019s Rule<\/h3>\n<hr \/>\n<p id=\"fs-id1165040721553\">Assume that [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right][\/latex]. Let <em data-effect=\"italics\">n<\/em> be a positive even integer and [latex]\\Delta x=\\frac{b-a}{n}[\/latex]. Let [latex]\\left[a,b\\right][\/latex] be divided into [latex]n[\/latex] subintervals, each of length [latex]\\Delta x[\/latex], with endpoints at [latex]P=\\left\\{{x}_{0},{x}_{1},{x}_{2},\\ldots ,{x}_{n}\\right\\}[\/latex]. Set<\/p>\n<div id=\"fs-id1165041953720\" style=\"text-align: center;\" data-type=\"equation\">[latex]{S}_{n}=\\frac{\\Delta x}{3}\\left(f\\left({x}_{0}\\right)+4f\\left({x}_{1}\\right)+2f\\left({x}_{2}\\right)+4f\\left({x}_{3}\\right)+2f\\left({x}_{4}\\right)+\\cdots +2f\\left({x}_{n - 2}\\right)+4f\\left({x}_{n - 1}\\right)+f\\left({x}_{n}\\right)\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042243662\">Then,<\/p>\n<div id=\"fs-id1165042243666\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\text{+}\\infty }{\\text{lim}}{S}_{n}={\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042108863\">Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson\u2019s rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that [latex]{S}_{2n}=\\left(\\frac{2}{3}\\right){M}_{n}+\\left(\\frac{1}{3}\\right){T}_{n}[\/latex].<\/p>\n<p id=\"fs-id1165042264461\">It is also possible to put a bound on the error when using Simpson\u2019s rule to approximate a definite integral. The bound in the error is given by the following rule:<\/p>\n<div id=\"fs-id1165042264468\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Rule: Error Bound for Simpson\u2019s Rule<\/h3>\n<hr \/>\n<p id=\"fs-id1165041836425\">Let [latex]f\\left(x\\right)[\/latex] be a continuous function over [latex]\\left[a,b\\right][\/latex] having a fourth derivative, [latex]{f}^{\\left(4\\right)}\\left(x\\right)[\/latex], over this interval. If [latex]M[\/latex] is the maximum value of [latex]|{f}^{\\left(4\\right)}\\left(x\\right)|[\/latex] over [latex]\\left[a,b\\right][\/latex], then the upper bound for the error in using [latex]{S}_{n}[\/latex] to estimate [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex] is given by<\/p>\n<div id=\"fs-id1165042109950\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\text{Error in}{S}_{n}\\le \\frac{M{\\left(b-a\\right)}^{5}}{180{n}^{4}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042217547\" data-type=\"example\">\n<div id=\"fs-id1165042217549\" data-type=\"exercise\">\n<div id=\"fs-id1165042217551\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Applying Simpson\u2019s Rule 1<\/h3>\n<div id=\"fs-id1165042217551\" data-type=\"problem\">\n<p id=\"fs-id1165042217557\">Use [latex]{S}_{2}[\/latex] to approximate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{3}dx[\/latex]. Estimate a bound for the error in [latex]{S}_{2}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558799\">Show Solution<\/span><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041848372\" data-type=\"solution\">\n<p id=\"fs-id1165041848374\">Since [latex]\\left[0,1\\right][\/latex] is divided into two intervals, each subinterval has length [latex]\\Delta x=\\frac{1 - 0}{2}=\\frac{1}{2}[\/latex]. The endpoints of these subintervals are [latex]\\left\\{0,\\frac{1}{2},1\\right\\}[\/latex]. If we set [latex]f\\left(x\\right)={x}^{3}[\/latex], then<\/p>\n<p id=\"fs-id1165041962775\">[latex]{S}_{4}=\\frac{1}{3}\\cdot \\frac{1}{2}\\left(f\\left(0\\right)+4f\\left(\\frac{1}{2}\\right)+f\\left(1\\right)\\right)=\\frac{1}{6}\\left(0+4\\cdot \\frac{1}{8}+1\\right)=\\frac{1}{4}[\/latex]. Since [latex]{f}^{\\left(4\\right)}\\left(x\\right)=0[\/latex] and consequently [latex]M=0[\/latex], we see that<\/p>\n<div id=\"fs-id1167793627790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Error in }{S}_{2}\\le \\frac{0{\\left(1\\right)}^{5}}{180\\cdot {2}^{4}}=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041971466\">This bound indicates that the value obtained through Simpson\u2019s rule is exact. A quick check will verify that, in fact, [latex]{\\displaystyle\\int }_{0}^{1}{x}^{3}dx=\\frac{1}{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042199249\" data-type=\"example\">\n<div id=\"fs-id1165042199251\" data-type=\"exercise\">\n<div id=\"fs-id1165042199253\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Applying Simpson\u2019s Rule 2<\/h3>\n<div id=\"fs-id1165042199253\" data-type=\"problem\">\n<p id=\"fs-id1165042199259\">Use [latex]{S}_{6}[\/latex] to estimate the length of the curve [latex]y=\\frac{1}{2}{x}^{2}[\/latex] over [latex]\\left[1,4\\right][\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558699\">Show Solution<\/span><\/p>\n<div id=\"q44558699\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040668581\" data-type=\"solution\">\n<p id=\"fs-id1165040668583\">The length of [latex]y=\\frac{1}{2}{x}^{2}[\/latex] over [latex]\\left[1,4\\right][\/latex] is [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{1+{x}^{2}}dx[\/latex]. If we divide [latex]\\left[1,4\\right][\/latex] into six subintervals, then each subinterval has length [latex]\\Delta x=\\frac{4 - 1}{6}=\\frac{1}{2}[\/latex], and the endpoints of the subintervals are [latex]\\left\\{1,\\frac{3}{2},2,\\frac{5}{2},3,\\frac{7}{2},4\\right\\}[\/latex]. Setting [latex]f\\left(x\\right)=\\sqrt{1+{x}^{2}}[\/latex],<\/p>\n<div id=\"fs-id1165040758286\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{6}=\\frac{1}{3}\\cdot \\frac{1}{2}\\left(f\\left(1\\right)+4f\\left(\\frac{3}{2}\\right)+2f\\left(2\\right)+4f\\left(\\frac{5}{2}\\right)+2f\\left(3\\right)+4f\\left(\\frac{7}{2}\\right)+f\\left(4\\right)\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042296653\">After substituting, we have<\/p>\n<div id=\"fs-id1165042296656\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {S}_{6}& =\\frac{1}{6}\\left(1.4142+4\\cdot 1.80278+2\\cdot 2.23607+4\\cdot 2.69258+2\\cdot 3.16228+4\\cdot 3.64005+4.12311\\right)\\hfill \\\\ & \\approx 8.14594.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example:\u00a0Applying Simpson\u2019s Rule 2<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XZ_mr5QqQNA?controls=0&amp;start=290&amp;end=543&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.6.5_290to543_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.6.5&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1165040796317\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165040796321\" data-type=\"exercise\">\n<div id=\"fs-id1165041899066\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165041899066\" data-type=\"problem\">\n<p id=\"fs-id1165041899068\">Use [latex]{S}_{2}[\/latex] to estimate [latex]{\\displaystyle\\int }_{1}^{2}\\frac{1}{x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558499\">Hint<\/span><\/p>\n<div id=\"q44558499\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040638592\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165042002716\">[latex]{S}_{2}=\\left(\\frac{1}{3}\\Delta x\\left(f\\left({x}_{0}\\right)+4f\\left({x}_{1}\\right)+f\\left({x}_{2}\\right)\\right)\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558599\">Show Solution<\/span><\/p>\n<div id=\"q44558599\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040638577\" data-type=\"solution\">\n<p id=\"fs-id1165040638579\">[latex]\\frac{25}{36}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165042276347\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5578\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5578&theme=oea&iframe_resize_id=ohm5578&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1599\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.6.5. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":22,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.6.5\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1599","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1599","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1599\/revisions"}],"predecessor-version":[{"id":2037,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1599\/revisions\/2037"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/158"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1599\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1599"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1599"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1599"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}