{"id":754,"date":"2025-06-20T17:10:11","date_gmt":"2025-06-20T17:10:11","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=754"},"modified":"2025-09-05T17:44:19","modified_gmt":"2025-09-05T17:44:19","slug":"numerical-integration-methods-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/numerical-integration-methods-fresh-take\/","title":{"raw":"Numerical Integration Methods: Fresh Take","rendered":"Numerical Integration Methods: Fresh Take"},"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\">The Midpoint Rule<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Many real-world functions don't have neat, closed-form antiderivatives that you can write down with elementary functions. When the Fundamental Theorem of Calculus hits a wall, numerical integration steps in to give you excellent approximations of definite integrals.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Midpoint Rule - Your First Tool: <\/strong>The midpoint rule uses rectangles with heights determined by function values at the midpoint of each subinterval. Here's the setup:<\/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>Divide the interval<\/strong> [latex][a,b][\/latex] into [latex]n[\/latex] equal subintervals, each with width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find the midpoint<\/strong> [latex]m_i[\/latex] of each subinterval<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Calculate the approximation:<\/strong> [latex]M_n = \\sum_{i=1}^n f(m_i) \\Delta x[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Midpoints often provide better approximations than left or right endpoints because they tend to balance out the over- and under-estimation errors, especially for curved functions.<\/p>\r\n<p class=\"whitespace-normal break-words\">More subintervals (larger [latex]n[\/latex]) generally mean better approximations, but they also mean more calculations. The key is finding the right balance between accuracy and computational effort.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042231132\" data-type=\"problem\">\r\n<p id=\"fs-id1165040706128\">Use the midpoint rule with [latex]n=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=\"44558896\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1165041834633\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165040743320\">[latex]\\Delta x=\\frac{1}{2}[\/latex], [latex]{m}_{1}=\\frac{5}{4}[\/latex], and [latex]{m}_{2}=\\frac{7}{4}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1165041815897\" data-type=\"solution\">\r\n<p id=\"fs-id1165041816976\">[latex]\\frac{24}{35}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">The Trapezoidal Rule<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The trapezoidal rule improves on basic Riemann sums by using trapezoids instead of rectangles. Instead of creating \"steps\" that miss curved portions, trapezoids connect adjacent function values with straight lines, following the curve's shape more naturally.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>How It Works:<\/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>Divide the interval<\/strong> [latex][a,b][\/latex] into [latex]n[\/latex] equal subintervals with width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Use the trapezoid area formula<\/strong> for each piece: [latex]\\frac{1}{2}\\Delta x(f(x_i) + f(x_{i+1}))[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Add up all trapezoid areas:<\/strong> [latex]T_n = \\frac{1}{2}\\Delta x(f(x_0) + 2f(x_1) + 2f(x_2) + \\cdots + 2f(x_{n-1}) + f(x_n))[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Notice that interior function values get multiplied by [latex]2[\/latex], while the endpoints [latex]f(x_0)[\/latex] and [latex]f(x_n)[\/latex] appear only once. This happens because interior points serve as endpoints for two adjacent trapezoids.<\/p>\r\n<p class=\"whitespace-normal break-words\">The trapezoidal rule is actually the average of left and right Riemann sums: [latex]T_n = \\frac{1}{2}(L_n + R_n)[\/latex]. This makes intuitive sense\u2014you're averaging the left and right endpoint heights for each subinterval.<\/p>\r\n<p class=\"whitespace-normal break-words\">Trapezoids provide better curve-following than rectangles, but the midpoint rule often gives even better approximations due to its error-balancing properties.<\/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\"><strong>Concave up functions<\/strong>: Trapezoids systematically overestimate (sit above the curve)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Concave down functions<\/strong>: Trapezoids systematically underestimate (sit below the curve)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Midpoint rule advantage<\/strong>: Often more accurate because errors tend to cancel out rather than accumulate in one direction<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165041814723\" data-type=\"problem\">\r\n<p id=\"fs-id1165041814725\">Use the trapezoidal rule with [latex]n=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=\"44558893\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1165040692039\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041803658\">Set [latex]\\Delta x=\\frac{1}{2}[\/latex]. The endpoints of the subintervals are the elements of the set [latex]P=\\left\\{1,\\frac{3}{2},2\\right\\}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1165041782503\" data-type=\"solution\">\r\n<p id=\"fs-id1165041782505\">[latex]\\frac{17}{24}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2>Simpson's Rule<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Simpson's rule takes numerical integration to the next level by approximating functions with parabolas instead of straight lines or rectangles. Since many real functions are curved, using quadratic approximations captures their shape much more accurately than simpler methods.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Setup Requirements:<\/strong><\/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\">Must use an <strong>even number<\/strong> of subintervals (this is non-negotiable!)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Each subinterval has equal width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Work in pairs: every two adjacent subintervals get fitted with one parabola<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>How It Works:<\/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>Take three consecutive points<\/strong> like [latex](x_0, f(x_0))[\/latex], [latex](x_1, f(x_1))[\/latex], [latex](x_2, f(x_2))[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Fit a unique parabola<\/strong> [latex]p(x) = Ax^2 + Bx + C[\/latex] through these points<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate the parabola<\/strong> over the two subintervals [latex][x_0, x_2][\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Repeat the process<\/strong> for the next three points, and so on<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]S_n = \\frac{\\Delta x}{3}(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + \\cdots + 4f(x_{n-1}) + f(x_n))[\/latex]<\/p>\r\n\r\n<\/div>\r\n<h2>Error Bound for Simpson\u2019s Rule<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Simpson's rule doesn't just give you an approximation\u2014it also comes with a mathematical guarantee about how accurate that approximation can be. The error bound formula tells you the maximum possible error, giving you confidence in your results.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Error Bound Formula:<\/strong> If [latex]M[\/latex] is the maximum value of [latex]|f^{(4)}(x)|[\/latex] over [latex][a,b][\/latex], then: [latex]\\text{Error in } S_n \\leq \\frac{M(b-a)^5}{180n^4}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Simpson's rule can be expressed as a weighted average: [latex]S_{2n} = \\frac{2}{3}M_n + \\frac{1}{3}T_n[\/latex] This shows how Simpson's rule cleverly combines the midpoint and trapezoidal rules.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Special Cases:<\/strong><\/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\">Polynomials of degree 3 or less: [latex]f^{(4)}(x) = 0[\/latex], so [latex]M = 0[\/latex] and the error is exactly zero<\/li>\r\n \t<li class=\"whitespace-normal break-words\">This means Simpson's rule gives exact answers for cubic and lower-degree polynomials<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/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>Find the fourth derivative<\/strong> of your function<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Determine its maximum absolute value<\/strong> over your interval<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Apply the error bound formula<\/strong> to see if your approximation meets your precision needs<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Adjust [latex]n[\/latex] if needed<\/strong> for better accuracy<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/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\">The Midpoint Rule<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Many real-world functions don&#8217;t have neat, closed-form antiderivatives that you can write down with elementary functions. When the Fundamental Theorem of Calculus hits a wall, numerical integration steps in to give you excellent approximations of definite integrals.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Midpoint Rule &#8211; Your First Tool: <\/strong>The midpoint rule uses rectangles with heights determined by function values at the midpoint of each subinterval. Here&#8217;s the setup:<\/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>Divide the interval<\/strong> [latex][a,b][\/latex] into [latex]n[\/latex] equal subintervals, each with width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find the midpoint<\/strong> [latex]m_i[\/latex] of each subinterval<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Calculate the approximation:<\/strong> [latex]M_n = \\sum_{i=1}^n f(m_i) \\Delta x[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Midpoints often provide better approximations than left or right endpoints because they tend to balance out the over- and under-estimation errors, especially for curved functions.<\/p>\n<p class=\"whitespace-normal break-words\">More subintervals (larger [latex]n[\/latex]) generally mean better approximations, but they also mean more calculations. The key is finding the right balance between accuracy and computational effort.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042231132\" data-type=\"problem\">\n<p id=\"fs-id1165040706128\">Use the midpoint rule with [latex]n=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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Hint<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041834633\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165040743320\">[latex]\\Delta x=\\frac{1}{2}[\/latex], [latex]{m}_{1}=\\frac{5}{4}[\/latex], and [latex]{m}_{2}=\\frac{7}{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Show Solution<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041815897\" data-type=\"solution\">\n<p id=\"fs-id1165041816976\">[latex]\\frac{24}{35}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">The Trapezoidal Rule<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The trapezoidal rule improves on basic Riemann sums by using trapezoids instead of rectangles. Instead of creating &#8220;steps&#8221; that miss curved portions, trapezoids connect adjacent function values with straight lines, following the curve&#8217;s shape more naturally.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>How It Works:<\/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>Divide the interval<\/strong> [latex][a,b][\/latex] into [latex]n[\/latex] equal subintervals with width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Use the trapezoid area formula<\/strong> for each piece: [latex]\\frac{1}{2}\\Delta x(f(x_i) + f(x_{i+1}))[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Add up all trapezoid areas:<\/strong> [latex]T_n = \\frac{1}{2}\\Delta x(f(x_0) + 2f(x_1) + 2f(x_2) + \\cdots + 2f(x_{n-1}) + f(x_n))[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Notice that interior function values get multiplied by [latex]2[\/latex], while the endpoints [latex]f(x_0)[\/latex] and [latex]f(x_n)[\/latex] appear only once. This happens because interior points serve as endpoints for two adjacent trapezoids.<\/p>\n<p class=\"whitespace-normal break-words\">The trapezoidal rule is actually the average of left and right Riemann sums: [latex]T_n = \\frac{1}{2}(L_n + R_n)[\/latex]. This makes intuitive sense\u2014you&#8217;re averaging the left and right endpoint heights for each subinterval.<\/p>\n<p class=\"whitespace-normal break-words\">Trapezoids provide better curve-following than rectangles, but the midpoint rule often gives even better approximations due to its error-balancing properties.<\/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\"><strong>Concave up functions<\/strong>: Trapezoids systematically overestimate (sit above the curve)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Concave down functions<\/strong>: Trapezoids systematically underestimate (sit below the curve)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Midpoint rule advantage<\/strong>: Often more accurate because errors tend to cancel out rather than accumulate in one direction<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165041814723\" data-type=\"problem\">\n<p id=\"fs-id1165041814725\">Use the trapezoidal rule with [latex]n=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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558893\">Hint<\/button><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040692039\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041803658\">Set [latex]\\Delta x=\\frac{1}{2}[\/latex]. The endpoints of the subintervals are the elements of the set [latex]P=\\left\\{1,\\frac{3}{2},2\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Show Solution<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041782503\" data-type=\"solution\">\n<p id=\"fs-id1165041782505\">[latex]\\frac{17}{24}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2>Simpson&#8217;s Rule<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Simpson&#8217;s rule takes numerical integration to the next level by approximating functions with parabolas instead of straight lines or rectangles. Since many real functions are curved, using quadratic approximations captures their shape much more accurately than simpler methods.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Setup Requirements:<\/strong><\/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\">Must use an <strong>even number<\/strong> of subintervals (this is non-negotiable!)<\/li>\n<li class=\"whitespace-normal break-words\">Each subinterval has equal width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Work in pairs: every two adjacent subintervals get fitted with one parabola<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>How It Works:<\/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>Take three consecutive points<\/strong> like [latex](x_0, f(x_0))[\/latex], [latex](x_1, f(x_1))[\/latex], [latex](x_2, f(x_2))[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Fit a unique parabola<\/strong> [latex]p(x) = Ax^2 + Bx + C[\/latex] through these points<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate the parabola<\/strong> over the two subintervals [latex][x_0, x_2][\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Repeat the process<\/strong> for the next three points, and so on<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]S_n = \\frac{\\Delta x}{3}(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + \\cdots + 4f(x_{n-1}) + f(x_n))[\/latex]<\/p>\n<\/div>\n<h2>Error Bound for Simpson\u2019s Rule<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Simpson&#8217;s rule doesn&#8217;t just give you an approximation\u2014it also comes with a mathematical guarantee about how accurate that approximation can be. The error bound formula tells you the maximum possible error, giving you confidence in your results.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Error Bound Formula:<\/strong> If [latex]M[\/latex] is the maximum value of [latex]|f^{(4)}(x)|[\/latex] over [latex][a,b][\/latex], then: [latex]\\text{Error in } S_n \\leq \\frac{M(b-a)^5}{180n^4}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Simpson&#8217;s rule can be expressed as a weighted average: [latex]S_{2n} = \\frac{2}{3}M_n + \\frac{1}{3}T_n[\/latex] This shows how Simpson&#8217;s rule cleverly combines the midpoint and trapezoidal rules.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Special Cases:<\/strong><\/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\">Polynomials of degree 3 or less: [latex]f^{(4)}(x) = 0[\/latex], so [latex]M = 0[\/latex] and the error is exactly zero<\/li>\n<li class=\"whitespace-normal break-words\">This means Simpson&#8217;s rule gives exact answers for cubic and lower-degree polynomials<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/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>Find the fourth derivative<\/strong> of your function<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Determine its maximum absolute value<\/strong> over your interval<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Apply the error bound formula<\/strong> to see if your approximation meets your precision needs<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Adjust [latex]n[\/latex] if needed<\/strong> for better accuracy<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558499\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558599\">Show Solution<\/button><\/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<\/section>\n","protected":false},"author":15,"menu_order":10,"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 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