{"id":751,"date":"2025-06-20T17:10:02","date_gmt":"2025-06-20T17:10:02","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=751"},"modified":"2025-07-18T16:07:08","modified_gmt":"2025-07-18T16:07:08","slug":"numerical-integration-methods-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/numerical-integration-methods-learn-it-3\/","title":{"raw":"Numerical Integration Methods: Learn It 3","rendered":"Numerical Integration Methods: Learn It 3"},"content":{"raw":"

Simpson's Rule<\/h2>\r\n

We have seen how the midpoint rule approximates curves with rectangles and the trapezoidal rule uses straight lines. Simpson's rule takes this progression further by approximating curves with parabolas<\/strong> (quadratic functions).<\/p>\r\n

This method uses piecewise quadratic functions to follow the curve more accurately than either rectangles or straight-line segments.<\/p>\r\n\r\n

How Simpson's Rule Works<\/h3>\r\n

The setup requires partitioning the interval into an even number<\/strong> of subintervals, each with equal width. This even number requirement is essential because we work with pairs of subintervals.<\/p>\r\n

For each pair of subintervals:<\/strong> Over the first pair of subintervals, we approximate [latex]\\int_{x_0}^{x_2} f(x)dx[\/latex] with [latex]\\int_{x_0}^{x_2} p(x)dx[\/latex], where [latex]p(x) = Ax^2 + Bx + C[\/latex] is the unique quadratic function passing through the three points [latex](x_0, f(x_0))[\/latex], [latex](x_1, f(x_1))[\/latex], and [latex](x_2, f(x_2))[\/latex].<\/p>\r\n

Continuing the pattern:<\/strong> Over the next pair of subintervals, we approximate [latex]\\int_{x_2}^{x_4} f(x)dx[\/latex] with the integral of another quadratic function passing through [latex](x_2, f(x_2))[\/latex], [latex](x_3, f(x_3))[\/latex], and [latex](x_4, f(x_4))[\/latex]. This process continues with each successive pair of subintervals.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"This Figure 4. With Simpson\u2019s rule, we approximate a definite integral by integrating a piecewise quadratic function.[\/caption]\r\n\r\n

Why Quadratics?<\/strong> Three points determine a unique parabola, just like two points determine a unique line. Using quadratic approximations captures the curvature of functions much better than straight-line approximations.<\/section>\r\n

Deriving Simpson's Rule Formula<\/h3>\r\n

To understand the formula for Simpson's rule, we need to derive the approximation over the first two subintervals. This derivation requires keeping track of several key relationships.<\/p>\r\n

The quadratic function [latex]p(x) = Ax^2 + Bx + C[\/latex] must pass through our three points:<\/p>\r\n

[latex]\\begin{array}{c}f(x_0) = p(x_0) = Ax_0^2 + Bx_0 + C\\\\ f(x_1) = p(x_1) = Ax_1^2 + Bx_1 + C\\\\ f(x_2) = p(x_2) = Ax_2^2 + Bx_2 + C\\end{array}[\/latex]<\/p>\r\n

We also need these geometric facts:<\/p>\r\n\r\n

    \r\n \t
  • [latex]x_2 - x_0 = 2\\Delta x[\/latex], where [latex]\\Delta x[\/latex] is the length of a subinterval<\/li>\r\n \t
  • [latex]x_2 + x_0 = 2x_1[\/latex], since [latex]x_1 = \\frac{(x_2 + x_0)}{2}[\/latex]<\/li>\r\n<\/ul>\r\n

    We can now integrate the quadratic approximation:<\/p>\r\n

    [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\\end{array}[\/latex]<\/p>\r\n

    After factoring and using algebraic identities:<\/p>\r\n

    [latex]\\begin{array}{ccccc} & =\\frac{A}{3}({x}_{2}-{x}_{0})({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2}) \u00a0+\\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\\end{array}[\/latex]<\/p>\r\n

    Through careful algebraic manipulation and substitution of our relationships, this eventually simplifies to:<\/p>\r\n

    [latex]\\int_{x_0}^{x_2} f(x)dx = \\frac{\\Delta x}{3}(f(x_0) + 4f(x_1) + f(x_2))[\/latex]<\/p>\r\n

    When we apply the same method to approximate [latex]\\int_{x_2}^{x_4} f(x)dx[\/latex], we get:<\/p>\r\n

    [latex]\\int_{x_2}^{x_4} f(x)dx \\approx \\frac{\\Delta x}{3}(f(x_2) + 4f(x_3) + f(x_4))[\/latex]<\/p>\r\n

    Combining these approximations over the interval [latex][x_0, x_4][\/latex]:<\/p>\r\n

    [latex]\\int_{x_0}^{x_4} f(x)dx = \\frac{\\Delta x}{3}(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4))[\/latex]<\/p>\r\n\r\n

    Notice the Pattern<\/strong>: The coefficients follow a specific pattern: 1, 4, 2, 4, 1. Interior points where parabolas meet get coefficient 2, while other interior points get coefficient 4<\/section>This pattern continues as we add pairs of subintervals, leading us to the general Simpson's rule formula.\r\n\r\n
    \r\n
    \r\n

    Simpson\u2019s Rule<\/h3>\r\nAssume that [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right][\/latex]. Let n 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].\r\n\r\nSet\r\n

    [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].<\/p>\r\nThen,\r\n

    [latex]\\underset{n\\to \\text{+}\\infty }{\\text{lim}}{S}_{n}={\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section>","rendered":"

    Simpson’s Rule<\/h2>\n

    We have seen how the midpoint rule approximates curves with rectangles and the trapezoidal rule uses straight lines. Simpson’s rule takes this progression further by approximating curves with parabolas<\/strong> (quadratic functions).<\/p>\n

    This method uses piecewise quadratic functions to follow the curve more accurately than either rectangles or straight-line segments.<\/p>\n

    How Simpson’s Rule Works<\/h3>\n

    The setup requires partitioning the interval into an even number<\/strong> of subintervals, each with equal width. This even number requirement is essential because we work with pairs of subintervals.<\/p>\n

    For each pair of subintervals:<\/strong> Over the first pair of subintervals, we approximate [latex]\\int_{x_0}^{x_2} f(x)dx[\/latex] with [latex]\\int_{x_0}^{x_2} p(x)dx[\/latex], where [latex]p(x) = Ax^2 + Bx + C[\/latex] is the unique quadratic function passing through the three points [latex](x_0, f(x_0))[\/latex], [latex](x_1, f(x_1))[\/latex], and [latex](x_2, f(x_2))[\/latex].<\/p>\n

    Continuing the pattern:<\/strong> Over the next pair of subintervals, we approximate [latex]\\int_{x_2}^{x_4} f(x)dx[\/latex] with the integral of another quadratic function passing through [latex](x_2, f(x_2))[\/latex], [latex](x_3, f(x_3))[\/latex], and [latex](x_4, f(x_4))[\/latex]. This process continues with each successive pair of subintervals.<\/p>\n

    \"This
    Figure 4. With Simpson\u2019s rule, we approximate a definite integral by integrating a piecewise quadratic function.<\/figcaption><\/figure>\n
    Why Quadratics?<\/strong> Three points determine a unique parabola, just like two points determine a unique line. Using quadratic approximations captures the curvature of functions much better than straight-line approximations.<\/section>\n

    Deriving Simpson’s Rule Formula<\/h3>\n

    To understand the formula for Simpson’s rule, we need to derive the approximation over the first two subintervals. This derivation requires keeping track of several key relationships.<\/p>\n

    The quadratic function [latex]p(x) = Ax^2 + Bx + C[\/latex] must pass through our three points:<\/p>\n

    [latex]\\begin{array}{c}f(x_0) = p(x_0) = Ax_0^2 + Bx_0 + C\\\\ f(x_1) = p(x_1) = Ax_1^2 + Bx_1 + C\\\\ f(x_2) = p(x_2) = Ax_2^2 + Bx_2 + C\\end{array}[\/latex]<\/p>\n

    We also need these geometric facts:<\/p>\n

      \n
    • [latex]x_2 - x_0 = 2\\Delta x[\/latex], where [latex]\\Delta x[\/latex] is the length of a subinterval<\/li>\n
    • [latex]x_2 + x_0 = 2x_1[\/latex], since [latex]x_1 = \\frac{(x_2 + x_0)}{2}[\/latex]<\/li>\n<\/ul>\n

      We can now integrate the quadratic approximation:<\/p>\n

      [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\\end{array}[\/latex]<\/p>\n

      After factoring and using algebraic identities:<\/p>\n

      [latex]\\begin{array}{ccccc} & =\\frac{A}{3}({x}_{2}-{x}_{0})({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2}) \u00a0+\\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\\end{array}[\/latex]<\/p>\n

      Through careful algebraic manipulation and substitution of our relationships, this eventually simplifies to:<\/p>\n

      [latex]\\int_{x_0}^{x_2} f(x)dx = \\frac{\\Delta x}{3}(f(x_0) + 4f(x_1) + f(x_2))[\/latex]<\/p>\n

      When we apply the same method to approximate [latex]\\int_{x_2}^{x_4} f(x)dx[\/latex], we get:<\/p>\n

      [latex]\\int_{x_2}^{x_4} f(x)dx \\approx \\frac{\\Delta x}{3}(f(x_2) + 4f(x_3) + f(x_4))[\/latex]<\/p>\n

      Combining these approximations over the interval [latex][x_0, x_4][\/latex]:<\/p>\n

      [latex]\\int_{x_0}^{x_4} f(x)dx = \\frac{\\Delta x}{3}(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4))[\/latex]<\/p>\n

      Notice the Pattern<\/strong>: The coefficients follow a specific pattern: 1, 4, 2, 4, 1. Interior points where parabolas meet get coefficient 2, while other interior points get coefficient 4<\/section>\n

      This pattern continues as we add pairs of subintervals, leading us to the general Simpson’s rule formula.<\/p>\n

      \n
      \n

      Simpson\u2019s Rule<\/h3>\n

      Assume that [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right][\/latex]. Let n 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].<\/p>\n

      Set<\/p>\n

      [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].<\/p>\n

      Then,<\/p>\n

      [latex]\\underset{n\\to \\text{+}\\infty }{\\text{lim}}{S}_{n}={\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex].<\/p>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":7,"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\/751"}],"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":14,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/751\/revisions"}],"predecessor-version":[{"id":1313,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/751\/revisions\/1313"}],"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\/751\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=751"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=751"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=751"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=751"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}