{"id":749,"date":"2025-06-20T17:09:57","date_gmt":"2025-06-20T17:09:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=749"},"modified":"2025-08-21T15:54:49","modified_gmt":"2025-08-21T15:54:49","slug":"numerical-integration-methods-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/numerical-integration-methods-learn-it-1\/","title":{"raw":"Numerical Integration Methods: Learn It 1","rendered":"Numerical Integration Methods: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Estimate definite integrals using the midpoint and trapezoidal rules<\/li>\r\n \t<li>Use Simpson's rule to find definite integrals with a specified accuracy<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Why Do We Need Numerical Integration?<\/h2>\r\n<p class=\"whitespace-normal break-words\">Here's the reality: many functions you'll encounter don't have antiderivatives that can be expressed in simple, closed forms. What does this mean for you? You can't always use the Fundamental Theorem of Calculus to evaluate definite integrals directly.<\/p>\r\n<p class=\"whitespace-normal break-words\">The solution? We use numerical integration techniques to approximate these integral values. Think of it as getting a really good estimate when an exact answer isn't practical.<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Engineers, scientists, and economists constantly use numerical integration. When modeling population growth, calculating areas under complex curves, or analyzing data that doesn't follow neat mathematical functions, these approximation methods are essential tools.<\/section>\r\n<p class=\"whitespace-normal break-words\">Remember that we defined definite integrals as limits of Riemann sums? Well, any Riemann sum can serve as an estimate for the integral [latex]\\displaystyle\\int_a^b f(x),dx[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Let's recall how Riemann sums work:<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Setting up a Riemann sum:<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Partition the interval<\/strong> [latex][a,b][\/latex] into subintervals: [latex]P = \\{x_0, x_1, x_2, \\ldots, x_n\\}[\/latex] where [latex]a = x_0 &lt; x_1 &lt; x_2 &lt; \\cdots &lt; x_n = b[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Choose sample points<\/strong> in each subinterval: [latex]S = \\{x_1^{*}, x_2^{*}, \\ldots, x_n^{*}\\}[\/latex] where [latex]x_{i-1} \\leq x_i^{*} \\leq x_i[\/latex] for all [latex]i[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Calculate the Riemann sum:<\/strong> [latex]\\sum_{i=1}^n f(x_i^*) \\Delta x_i[\/latex] where [latex]\\Delta x_i = x_i - x_{i-1}[\/latex]<\/li>\r\n<\/ol>\r\nThe width [latex]\\Delta x_i[\/latex] represents the length of the [latex]i[\/latex]th subinterval, and [latex]f(x_i^*)[\/latex] gives us the height of each rectangle in our approximation.\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">The Midpoint Rule<\/h2>\r\n<p class=\"whitespace-normal break-words\">One specific and very useful type of Riemann sum is the <strong>midpoint rule<\/strong>. Here's what makes it special:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Use subintervals of <strong>equal width<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">Choose the <strong>midpoint<\/strong> [latex]m_i[\/latex] of each subinterval as your sample point [latex]x_i^*[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">This approach often gives you better approximations than using left or right endpoints, especially when the function is curved.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<div>\r\n<h3>The Midpoint Rule<\/h3>\r\nAssume that [latex]f\\left(x\\right)[\/latex] is continuous on [latex]\\left[a,b\\right][\/latex]. Let n be a positive integer and [latex]\\Delta x=\\frac{b-a}{n}[\/latex]. If [latex]\\left[a,b\\right][\/latex] is divided into [latex]n[\/latex] subintervals, each of length [latex]\\Delta x[\/latex], and [latex]{m}_{i}[\/latex] is the midpoint of the [latex]i[\/latex]th subinterval, set\r\n\r\n<center>[latex]{M}_{n}=\\displaystyle\\sum _{i=1}^{n}f\\left({m}_{i}\\right)\\Delta x[\/latex].<\/center>Then [latex]\\underset{n\\to \\infty }{\\text{lim}}{M}_{n}={\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex].\r\n\r\n<\/div>\r\n<\/section>\r\n<p id=\"fs-id1165041977672\">As we can see in Figure 1, if [latex]f\\left(x\\right)\\ge 0[\/latex] over [latex]\\left[a,b\\right][\/latex], then [latex]\\displaystyle\\sum _{i=1}^{n}f\\left({m}_{i}\\right)\\Delta x[\/latex] corresponds to the sum of the areas of rectangles approximating the area between the graph of [latex]f\\left(x\\right)[\/latex] and the [latex]x[\/latex]-axis over [latex]\\left[a,b\\right][\/latex]. The graph shows the rectangles corresponding to [latex]{M}_{4}[\/latex] for a nonnegative function over a closed interval [latex]\\left[a,b\\right][\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_06_001\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233831\/CNX_Calc_Figure_07_06_001.jpg\" alt=\"This figure is a graph of a non-negative function in the first quadrant. The function increases and decreases. The quadrant is divided into a grid. Beginning on the x-axis at the point labeled a = x sub 0, there are rectangles shaded whose heights are approximately the height of the curve. The x-axis is scaled by increments of msub1, x sub 1, m sub 2, x sub 2, m sub 3, x sub 3, m sub 4 and b = x sub 4.\" width=\"487\" height=\"286\" data-media-type=\"image\/jpeg\" \/> Figure 1. The midpoint rule approximates the area between the graph of [latex]f\\left(x\\right)[\/latex] and the x-axis by summing the areas of rectangles with midpoints that are points on [latex]f\\left(x\\right)[\/latex].[\/caption]<\/figure>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165041843774\" data-type=\"problem\">\r\n<p id=\"fs-id1165040716416\">Use the midpoint rule to estimate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}dx[\/latex] using four subintervals. Compare the result with the actual value of this integral.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1165042276048\" data-type=\"solution\">\r\n<p id=\"fs-id1165042050773\">Each subinterval has length [latex]\\Delta x=\\frac{1 - 0}{4}=\\frac{1}{4}[\/latex]. Therefore, the subintervals consist of<\/p>\r\n\r\n<div id=\"fs-id1165042066546\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left[0,\\frac{1}{4}\\right],\\left[\\frac{1}{4},\\frac{1}{2}\\right],\\left[\\frac{1}{2},\\frac{3}{4}\\right],\\text{and}\\left[\\frac{3}{4},1\\right][\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041815762\">The midpoints of these subintervals are [latex]\\left\\{\\frac{1}{8},\\frac{3}{8},\\frac{5}{8},\\frac{7}{8}\\right\\}[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165041791048\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{M}_{4}=\\frac{1}{4}f\\left(\\frac{1}{8}\\right)+\\frac{1}{4}f\\left(\\frac{3}{8}\\right)+\\frac{1}{4}f\\left(\\frac{5}{8}\\right)+\\frac{1}{4}f\\left(\\frac{7}{8}\\right)=\\frac{1}{4}\\cdot \\frac{1}{64}+\\frac{1}{4}\\cdot \\frac{9}{64}+\\frac{1}{4}\\cdot \\frac{25}{64}+\\frac{1}{4}\\cdot \\frac{21}{64}=\\frac{21}{64}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041770066\">Since<\/p>\r\n\r\n<div id=\"fs-id1165041915328\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}dx=\\frac{1}{3}\\text{and}|\\frac{1}{3}-\\frac{21}{64}|=\\frac{1}{192}\\approx 0.0052[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042134377\">we see that the midpoint rule produces an estimate that is somewhat close to the actual value of the definite integral.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165041827477\" data-type=\"problem\">\r\n<p id=\"fs-id1165042128953\">Use [latex]{M}_{6}[\/latex] to estimate the length of the curve [latex]y=\\frac{1}{2}{x}^{2}[\/latex] on [latex]\\left[1,4\\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1165040754773\" data-type=\"solution\">\r\n<p id=\"fs-id1165041805251\">The length of [latex]y=\\frac{1}{2}{x}^{2}[\/latex] on [latex]\\left[1,4\\right][\/latex] is<\/p>\r\n\r\n<div id=\"fs-id1165041813602\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{1+{\\left(\\frac{dy}{dx}\\right)}^{2}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040691246\">Since [latex]\\frac{dy}{dx}=x[\/latex], this integral becomes [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{1+{x}^{2}}dx[\/latex].<\/p>\r\n<p id=\"fs-id1165041893161\">If [latex]\\left[1,4\\right][\/latex] is divided into six subintervals, then each subinterval has length [latex]\\Delta x=\\frac{4 - 1}{6}=\\frac{1}{2}[\/latex] and the midpoints of the subintervals are [latex]\\left\\{\\frac{5}{4},\\frac{7}{4},\\frac{9}{4},\\frac{11}{4},\\frac{13}{4},\\frac{15}{4}\\right\\}[\/latex]. If we set [latex]f\\left(x\\right)=\\sqrt{1+{x}^{2}}[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165042017909\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {M}_{6}&amp; =\\frac{1}{2}f\\left(\\frac{5}{4}\\right)+\\frac{1}{2}f\\left(\\frac{7}{4}\\right)+\\frac{1}{2}f\\left(\\frac{9}{4}\\right)+\\frac{1}{2}f\\left(\\frac{11}{4}\\right)+\\frac{1}{2}f\\left(\\frac{13}{4}\\right)+\\frac{1}{2}f\\left(\\frac{15}{4}\\right)\\hfill \\\\ &amp; \\approx \\frac{1}{2}\\left(1.6008+2.0156+2.4622+2.9262+3.4004+3.8810\\right)=8.1431.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Estimate definite integrals using the midpoint and trapezoidal rules<\/li>\n<li>Use Simpson&#8217;s rule to find definite integrals with a specified accuracy<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Why Do We Need Numerical Integration?<\/h2>\n<p class=\"whitespace-normal break-words\">Here&#8217;s the reality: many functions you&#8217;ll encounter don&#8217;t have antiderivatives that can be expressed in simple, closed forms. What does this mean for you? You can&#8217;t always use the Fundamental Theorem of Calculus to evaluate definite integrals directly.<\/p>\n<p class=\"whitespace-normal break-words\">The solution? We use numerical integration techniques to approximate these integral values. Think of it as getting a really good estimate when an exact answer isn&#8217;t practical.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Engineers, scientists, and economists constantly use numerical integration. When modeling population growth, calculating areas under complex curves, or analyzing data that doesn&#8217;t follow neat mathematical functions, these approximation methods are essential tools.<\/section>\n<p class=\"whitespace-normal break-words\">Remember that we defined definite integrals as limits of Riemann sums? Well, any Riemann sum can serve as an estimate for the integral [latex]\\displaystyle\\int_a^b f(x),dx[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Let&#8217;s recall how Riemann sums work:<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Setting up a Riemann sum:<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Partition the interval<\/strong> [latex][a,b][\/latex] into subintervals: [latex]P = \\{x_0, x_1, x_2, \\ldots, x_n\\}[\/latex] where [latex]a = x_0 < x_1 < x_2 < \\cdots < x_n = b[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Choose sample points<\/strong> in each subinterval: [latex]S = \\{x_1^{*}, x_2^{*}, \\ldots, x_n^{*}\\}[\/latex] where [latex]x_{i-1} \\leq x_i^{*} \\leq x_i[\/latex] for all [latex]i[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Calculate the Riemann sum:<\/strong> [latex]\\sum_{i=1}^n f(x_i^*) \\Delta x_i[\/latex] where [latex]\\Delta x_i = x_i - x_{i-1}[\/latex]<\/li>\n<\/ol>\n<p>The width [latex]\\Delta x_i[\/latex] represents the length of the [latex]i[\/latex]th subinterval, and [latex]f(x_i^*)[\/latex] gives us the height of each rectangle in our approximation.<\/p>\n<\/section>\n<h2 data-type=\"title\">The Midpoint Rule<\/h2>\n<p class=\"whitespace-normal break-words\">One specific and very useful type of Riemann sum is the <strong>midpoint rule<\/strong>. Here&#8217;s what makes it special:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Use subintervals of <strong>equal width<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">Choose the <strong>midpoint<\/strong> [latex]m_i[\/latex] of each subinterval as your sample point [latex]x_i^*[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">This approach often gives you better approximations than using left or right endpoints, especially when the function is curved.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<div>\n<h3>The Midpoint Rule<\/h3>\n<p>Assume that [latex]f\\left(x\\right)[\/latex] is continuous on [latex]\\left[a,b\\right][\/latex]. Let n be a positive integer and [latex]\\Delta x=\\frac{b-a}{n}[\/latex]. If [latex]\\left[a,b\\right][\/latex] is divided into [latex]n[\/latex] subintervals, each of length [latex]\\Delta x[\/latex], and [latex]{m}_{i}[\/latex] is the midpoint of the [latex]i[\/latex]th subinterval, set<\/p>\n<div style=\"text-align: center;\">[latex]{M}_{n}=\\displaystyle\\sum _{i=1}^{n}f\\left({m}_{i}\\right)\\Delta x[\/latex].<\/div>\n<p>Then [latex]\\underset{n\\to \\infty }{\\text{lim}}{M}_{n}={\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex].<\/p>\n<\/div>\n<\/section>\n<p id=\"fs-id1165041977672\">As we can see in Figure 1, if [latex]f\\left(x\\right)\\ge 0[\/latex] over [latex]\\left[a,b\\right][\/latex], then [latex]\\displaystyle\\sum _{i=1}^{n}f\\left({m}_{i}\\right)\\Delta x[\/latex] corresponds to the sum of the areas of rectangles approximating the area between the graph of [latex]f\\left(x\\right)[\/latex] and the [latex]x[\/latex]-axis over [latex]\\left[a,b\\right][\/latex]. The graph shows the rectangles corresponding to [latex]{M}_{4}[\/latex] for a nonnegative function over a closed interval [latex]\\left[a,b\\right][\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_07_06_001\">\n<figure style=\"width: 487px\" 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\/11233831\/CNX_Calc_Figure_07_06_001.jpg\" alt=\"This figure is a graph of a non-negative function in the first quadrant. The function increases and decreases. The quadrant is divided into a grid. Beginning on the x-axis at the point labeled a = x sub 0, there are rectangles shaded whose heights are approximately the height of the curve. The x-axis is scaled by increments of msub1, x sub 1, m sub 2, x sub 2, m sub 3, x sub 3, m sub 4 and b = x sub 4.\" width=\"487\" height=\"286\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. The midpoint rule approximates the area between the graph of [latex]f\\left(x\\right)[\/latex] and the x-axis by summing the areas of rectangles with midpoints that are points on [latex]f\\left(x\\right)[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165041843774\" data-type=\"problem\">\n<p id=\"fs-id1165040716416\">Use the midpoint rule to estimate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}dx[\/latex] using four subintervals. Compare the result with the actual value of this integral.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042276048\" data-type=\"solution\">\n<p id=\"fs-id1165042050773\">Each subinterval has length [latex]\\Delta x=\\frac{1 - 0}{4}=\\frac{1}{4}[\/latex]. Therefore, the subintervals consist of<\/p>\n<div id=\"fs-id1165042066546\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left[0,\\frac{1}{4}\\right],\\left[\\frac{1}{4},\\frac{1}{2}\\right],\\left[\\frac{1}{2},\\frac{3}{4}\\right],\\text{and}\\left[\\frac{3}{4},1\\right][\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041815762\">The midpoints of these subintervals are [latex]\\left\\{\\frac{1}{8},\\frac{3}{8},\\frac{5}{8},\\frac{7}{8}\\right\\}[\/latex]. Thus,<\/p>\n<div id=\"fs-id1165041791048\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{M}_{4}=\\frac{1}{4}f\\left(\\frac{1}{8}\\right)+\\frac{1}{4}f\\left(\\frac{3}{8}\\right)+\\frac{1}{4}f\\left(\\frac{5}{8}\\right)+\\frac{1}{4}f\\left(\\frac{7}{8}\\right)=\\frac{1}{4}\\cdot \\frac{1}{64}+\\frac{1}{4}\\cdot \\frac{9}{64}+\\frac{1}{4}\\cdot \\frac{25}{64}+\\frac{1}{4}\\cdot \\frac{21}{64}=\\frac{21}{64}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041770066\">Since<\/p>\n<div id=\"fs-id1165041915328\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}dx=\\frac{1}{3}\\text{and}|\\frac{1}{3}-\\frac{21}{64}|=\\frac{1}{192}\\approx 0.0052[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042134377\">we see that the midpoint rule produces an estimate that is somewhat close to the actual value of the definite integral.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165041827477\" data-type=\"problem\">\n<p id=\"fs-id1165042128953\">Use [latex]{M}_{6}[\/latex] to estimate the length of the curve [latex]y=\\frac{1}{2}{x}^{2}[\/latex] on [latex]\\left[1,4\\right][\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040754773\" data-type=\"solution\">\n<p id=\"fs-id1165041805251\">The length of [latex]y=\\frac{1}{2}{x}^{2}[\/latex] on [latex]\\left[1,4\\right][\/latex] is<\/p>\n<div id=\"fs-id1165041813602\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{1+{\\left(\\frac{dy}{dx}\\right)}^{2}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040691246\">Since [latex]\\frac{dy}{dx}=x[\/latex], this integral becomes [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{1+{x}^{2}}dx[\/latex].<\/p>\n<p id=\"fs-id1165041893161\">If [latex]\\left[1,4\\right][\/latex] is divided into six subintervals, then each subinterval has length [latex]\\Delta x=\\frac{4 - 1}{6}=\\frac{1}{2}[\/latex] and the midpoints of the subintervals are [latex]\\left\\{\\frac{5}{4},\\frac{7}{4},\\frac{9}{4},\\frac{11}{4},\\frac{13}{4},\\frac{15}{4}\\right\\}[\/latex]. If we set [latex]f\\left(x\\right)=\\sqrt{1+{x}^{2}}[\/latex],<\/p>\n<div id=\"fs-id1165042017909\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {M}_{6}& =\\frac{1}{2}f\\left(\\frac{5}{4}\\right)+\\frac{1}{2}f\\left(\\frac{7}{4}\\right)+\\frac{1}{2}f\\left(\\frac{9}{4}\\right)+\\frac{1}{2}f\\left(\\frac{11}{4}\\right)+\\frac{1}{2}f\\left(\\frac{13}{4}\\right)+\\frac{1}{2}f\\left(\\frac{15}{4}\\right)\\hfill \\\\ & \\approx \\frac{1}{2}\\left(1.6008+2.0156+2.4622+2.9262+3.4004+3.8810\\right)=8.1431.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":667,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/749"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/749\/revisions"}],"predecessor-version":[{"id":1954,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/749\/revisions\/1954"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/667"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/749\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=749"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=749"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=749"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=749"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}