{"id":378,"date":"2025-02-13T19:44:27","date_gmt":"2025-02-13T19:44:27","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-areas-learn-it-3\/"},"modified":"2025-02-13T19:44:27","modified_gmt":"2025-02-13T19:44:27","slug":"approximating-areas-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-areas-learn-it-3\/","title":{"raw":"Approximating Areas: Learn It 3","rendered":"Approximating Areas: Learn It 3"},"content":{"raw":"\n

Riemann Sums<\/h2>\n
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So far we have been using rectangles to approximate the area under a curve. We've determined the heights of these rectangles by evaluating the function at either the right or left endpoints of each subinterval [latex][x_{i-1},x_i][\/latex].<\/p>\n

However, we do not have to restrict the evaluation to just these points. We can evaluate the function at any point [latex]x_i^*[\/latex] within the subinterval [latex][x_{i-1},x_i][\/latex], and use [latex]f(x_i^*)[\/latex] as the height of our rectangle. This gives us an estimate for the area of the form:<\/p>\n

[latex]A \\approx \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n

A sum of this form is called a Riemann sum<\/strong>, named after the 19th-century mathematician Bernhard Riemann.<\/p>\n

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Riemann sum<\/h3>\n\nLet [latex]f(x)[\/latex] be defined on a closed interval [latex][a,b][\/latex] and let [latex]P[\/latex] be a regular partition of [latex][a,b][\/latex]. Let [latex]\\Delta x[\/latex] be the width of each subinterval [latex][x_{i-1},x_i][\/latex] and for each [latex]i[\/latex], let [latex]x_i^*[\/latex] be any point in [latex][x_{i-1},x_i][\/latex].<\/div>\n
 <\/div>\n
A Riemann sum is defined for [latex]f(x)[\/latex] as<\/div>\n
[latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n<\/section>\n

When using left- and right-endpoint approximations, our estimates improve as we increase the number of subintervals [latex]n[\/latex]. The same idea applies to Riemann sums: the more subintervals we use, the better our approximation. Now, let's define the area under a curve using Riemann sums.<\/p>\n

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area under a curve using Riemann sums<\/h3>\n\nLet [latex]f(x)[\/latex] be a continuous, nonnegative function on an interval [latex][a,b][\/latex], and let [latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex] be a Riemann sum for [latex]f(x)[\/latex].<\/div>\n
 <\/div>\n
Then, the area under the curve [latex]y=f(x)[\/latex] on [latex][a,b][\/latex] is given by<\/div>\n
[latex]A=\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n<\/section>\n

Taking the limit of a sum is a bit different from taking the limit of a function [latex]f(x)[\/latex] as [latex]x[\/latex] goes to infinity. We discuss limits of sums in more detail in the chapter on Sequences and Series in Calculus 2. For now, assume that the techniques we use to compute limits of functions also apply to sums.<\/p>\n

We must also consider what happens if our sum converges to different limits for different choices of [latex]x_i^*[\/latex]. If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], the limit:<\/p>\n

[latex]\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex]<\/p>\n

is unique and does not depend on the choice of [latex]x_i^*[\/latex].<\/p>\n

Before we dive into examples, let's discuss some specific choices for [latex]x_i^*[\/latex]. Any choice for [latex]x_i^*[\/latex] gives us an estimate of the area under the curve, but we might want to know if our estimate is too high or too low. We can choose [latex]x_i^*[\/latex] to guarantee one result or the other.<\/p>\n

    \n\t
  • Overestimate:<\/strong> Choose [latex]x_i^*[\/latex] so that [latex]f(x_i^*)[\/latex] is the maximum value on [latex][x_{i-1},x_i][\/latex]. This makes our Riemann sum an upper sum<\/strong>.<\/li>\n\t
  • Underestimate:<\/strong> Choose [latex]x_i^*[\/latex] so that [latex]f(x_i^*)[\/latex] is the minimum value on [latex][x_{i-1},x_i][\/latex]. This makes our Riemann sum a lower sum<\/strong>.<\/li>\n<\/ul>\n
    \n
      \n\t
    • If a function is increasing over an interval, using the right endpoints for the upper sum and the left endpoints for the lower sum gives us a good estimate.<\/li>\n\t
    • If a function is decreasing, using the left endpoints for the upper sum and the right endpoints for the lower sum is effective.<\/li>\n<\/ul>\n<\/section>\n
      \n

      Find a lower sum for [latex]f(x)=10-x^2[\/latex] on [latex][1,2][\/latex]; use [latex]n=4[\/latex] subintervals.<\/p>\n

      [reveal-answer q=\"448179\"]Show Solution[\/reveal-answer]
      \n[hidden-answer a=\"448179\"]With [latex]n=4[\/latex] over the interval [latex][1,2], \\, \\Delta x=\\frac{1}{4}[\/latex]. We can list the intervals as [latex][1,1.25], \\, [1.25,1.5], \\, [1.5,1.75], \\, [1.75,2][\/latex]. Because the function is decreasing over the interval [latex][1,2][\/latex], (Figure 14) shows that a lower sum is obtained by using the right endpoints.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"The Figure 14. The graph of [latex]f(x)=10-x^2[\/latex] is set up for a right-endpoint approximation of the area bounded by the curve and the x-axis on [latex][1,2][\/latex], and it shows a lower sum.[\/caption]\n\n

      The Riemann sum is<\/p>\n

      [latex]\\begin{array}{ll}\\displaystyle\\sum_{k=1}^{4} (10-x^2)(0.25)& =0.25[10-(1.25)^2+10-(1.5)^2+10-(1.75)^2+10-(2)^2] \\\\ & =0.25[8.4375+7.75+6.9375+6] \\\\ & =7.28 \\end{array}[\/latex]<\/div>\n

       <\/p>\n

      The area of [latex]7.28[\/latex] is a lower sum and an underestimate.<\/p>\n

      Watch the following video to see the worked solution to this example.<\/p>\n