{"id":750,"date":"2025-06-20T17:10:00","date_gmt":"2025-06-20T17:10:00","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=750"},"modified":"2025-09-10T14:29:44","modified_gmt":"2025-09-10T14:29:44","slug":"numerical-integration-methods-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/numerical-integration-methods-learn-it-2\/","title":{"raw":"Numerical Integration Methods: Learn It 2","rendered":"Numerical Integration Methods: Learn It 2"},"content":{"raw":"

The Trapezoidal Rule<\/h2>\r\nSometimes rectangles aren't the best shape for approximating the area under a curve. What if we used trapezoids instead? This often gives us better approximations, especially when the function has significant curvature.\r\n\r\n
Why do trapezoids work better?<\/strong>\u00a0Rectangles create \"steps\" that miss curved portions of the function. Trapezoids follow the curve more closely by connecting function values with straight lines.<\/section>In Figure 2, the area beneath the curve is approximated by trapezoids rather than by rectangles.\r\n
<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 2. Trapezoids may be used to approximate the area under a curve, hence approximating the definite integral.[\/caption]<\/figure>\r\n

How the Trapezoidal Rule Works<\/h3>\r\n

The trapezoidal rule approximates the area under a curve by dividing the region into trapezoids instead of rectangles. Each subinterval has equal width [latex]\\Delta x[\/latex], and the \"height\" of each trapezoid is [latex]\\Delta x[\/latex] (measured horizontally). The parallel bases have lengths [latex]f(x_i)[\/latex] and [latex]f(x_{i+1})[\/latex].<\/p>\r\n\r\n

You'll need the trapezoid area formula to calculate each piece:For a trapezoid with height [latex]h[\/latex] and parallel bases of lengths [latex]b_1[\/latex] and [latex]b_2[\/latex], the area is [latex]\\text{Area} = \\frac{1}{2}h(b_1 + b_2)[\/latex].<\/section>\r\n

Let's work through how we get the trapezoidal rule formula. Consider the first few trapezoids:<\/p>\r\n\r\n

    \r\n \t
  • First trapezoid area:<\/strong> [latex]\\frac{1}{2}\\Delta x(f(x_0) + f(x_1))[\/latex]<\/li>\r\n \t
  • Second trapezoid area:<\/strong> [latex]\\frac{1}{2}\\Delta x(f(x_1) + f(x_2))[\/latex]<\/li>\r\n \t
  • Third trapezoid area:<\/strong> [latex]\\frac{1}{2}\\Delta x(f(x_2) + f(x_3))[\/latex]<\/li>\r\n \t
  • Fourth trapezoid area:<\/strong> [latex]\\frac{1}{2}\\Delta x(f(x_3) + f(x_4))[\/latex]<\/li>\r\n<\/ul>\r\n

    When we add up all the trapezoid areas:<\/p>\r\n

    [latex]\\int_a^b f(x),dx \\approx \\frac{1}{2}\\Delta x(f(x_0) + f(x_1)) + \\frac{1}{2}\\Delta x(f(x_1) + f(x_2)) + \\frac{1}{2}\\Delta x(f(x_2) + f(x_3)) + \\frac{1}{2}\\Delta x(f(x_3) + f(x_4))[\/latex]<\/p>\r\n

    Now here's the key insight: factor out [latex]\\frac{1}{2}\\Delta x[\/latex] and combine like terms:<\/p>\r\n

    [latex]\\int_a^b f(x),dx \\approx \\frac{1}{2}\\Delta x(f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4))[\/latex]<\/p>\r\n\r\n

    Notice that the first and last function values appear once, while all the interior values are multiplied by 2. This pattern holds no matter how many subintervals you use!<\/section>This approach generalizes to give us the formal trapezoidal rule.\r\n\r\n
    \r\n
    \r\n

    the trapezoidal 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 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\r\n\r\n
    [latex]{T}_{n}=\\frac{1}{2}\\Delta x\\left(f\\left({x}_{0}\\right)+2f\\left({x}_{1}\\right)+2f\\left({x}_{2}\\right)+\\cdots +2f\\left({x}_{n - 1}\\right)+f\\left({x}_{n}\\right)\\right)[\/latex].<\/center>Then, [latex]\\underset{n\\to \\text{+}\\infty }{\\text{lim}}{T}_{n}={\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex].\r\n\r\n<\/div>\r\n<\/section>\r\n

    Before we move forward, let's unpack some important patterns you should notice about how the trapezoidal rule works.<\/p>\r\n

    The trapezoidal rule is actually an average: <\/strong>the trapezoidal rule [latex]T_n[\/latex] is exactly the average of the left-hand and right-hand Riemann sums! Mathematically, this means:<\/p>\r\n

    [latex]T_n = \\frac{1}{2}(L_n + R_n)[\/latex]<\/p>\r\n

    where [latex]L_n = \\sum_{i=1}^n f(x_{i-1})\\Delta x[\/latex] and [latex]R_n = \\sum_{i=1}^n f(x_i)\\Delta x[\/latex]<\/p>\r\n

    This makes sense when you think about it\u2014each trapezoid connects the left endpoint height to the right endpoint height with a straight line.<\/p>\r\n

    The shape of your function determines whether you'll get an overestimate or underestimate:<\/p>\r\n\r\n

      \r\n \t
    • Trapezoidal rule behavior:<\/strong>\r\n
        \r\n \t
      • Concave up functions<\/strong>: The rule systematically overestimates the integral (trapezoids sit above the curve)<\/li>\r\n \t
      • Concave down functions<\/strong>: The rule systematically underestimates the integral (trapezoids sit below the curve)<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
      • Midpoint rule behavior:<\/strong> The midpoint rule is often more balanced. It tends to partially overestimate and partially underestimate over the same intervals, so these errors somewhat cancel each other out.<\/li>\r\n<\/ul>\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"This Figure 3. The trapezoidal rule tends to be less accurate than the midpoint rule.[\/caption]\r\n\r\n
        In general, you can expect the midpoint rule to be more accurate than the trapezoidal rule for most functions. The midpoint rule's error-canceling behavior gives it an edge!<\/section>
        \r\n
        \r\n

        Use the trapezoidal rule to estimate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}dx[\/latex] using four subintervals.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n

        \r\n

        The endpoints of the subintervals consist of elements of the set [latex]P=\\left\\{0,\\frac{1}{4},\\frac{1}{2},\\frac{3}{4},1\\right\\}[\/latex] and [latex]\\Delta x=\\frac{1 - 0}{4}=\\frac{1}{4}[\/latex]. Thus,<\/p>\r\n\r\n

        [latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int }_{0}^{1}{x}^{2}dx& \\approx \\frac{1}{2}\\cdot \\frac{1}{4}\\left(f\\left(0\\right)+2f\\left(\\frac{1}{4}\\right)+2f\\left(\\frac{1}{2}\\right)+2f\\left(\\frac{3}{4}\\right)+f\\left(1\\right)\\right)\\hfill \\\\ & =\\frac{1}{8}\\left(0+2\\cdot \\frac{1}{16}+2\\cdot \\frac{1}{4}+2\\cdot \\frac{9}{16}+1\\right)\\hfill \\\\ & =\\frac{11}{32}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        Watch the following video to see the worked solution to the example above.