{"id":1118,"date":"2021-06-30T17:01:55","date_gmt":"2021-06-30T17:01:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/riemann-sums\/"},"modified":"2022-03-19T03:11:33","modified_gmt":"2022-03-19T03:11:33","slug":"riemann-sums","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/riemann-sums\/","title":{"raw":"Riemann Sums","rendered":"Riemann Sums"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use Riemann sums to approximate area<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1170572551818\" class=\"bc-section section\">\r\n<p id=\"fs-id1170572551824\">So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval [latex][x_{i-1},x_i][\/latex]. In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point [latex]x_i^*[\/latex]\u00a0in 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>\r\n\r\n<div id=\"fs-id1170572376449\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A \\approx \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572376498\">A sum of this form is called a<strong> Riemann sum<\/strong>, named for the 19th-century mathematician Bernhard Riemann, who developed the idea.<\/p>\r\n\r\n<div id=\"fs-id1170572376504\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572376508\">Let [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]. A <strong>Riemann sum<\/strong> is defined for [latex]f(x)[\/latex] as<\/p>\r\n\r\n<div id=\"fs-id1170572444290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572344248\">Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as [latex]n[\/latex] get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of [latex]n[\/latex]. We are now ready to define the area under a curve in terms of Riemann sums.<\/p>\r\n\r\n<div id=\"fs-id1170572344262\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572344265\">Let [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]. Then, the <strong>area under the curve<\/strong> [latex]y=f(x)[\/latex] on [latex][a,b][\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1170572229819\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572217482\" class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<a href=\"http:\/\/archives.math.utk.edu\/visual.calculus\/4\/riemann_sums.4\/\" target=\"_blank\" rel=\"noopener\">See a graphical demonstration of the construction of a Riemann sum.<\/a>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572217494\">Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit of a function [latex]f(x)[\/latex] as [latex]x[\/latex] goes to infinity. Limits of sums are discussed in detail in the chapter on <a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/HTmjSAcf@2.46:7JGyeadv@3\/Introduction\">Sequences and Series<\/a>\u00a0in the second volume of this text; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums.<\/p>\r\n<p id=\"fs-id1170572217522\">Second, we must consider what to do if the expression converges to different limits for different choices of [latex]\\{x_i^*\\}[\/latex]. Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if [latex]f(x)[\/latex] is continuous on the closed interval [latex][a,b][\/latex], then [latex]\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex] exists and is unique (in other words, it does not depend on the choice of [latex]\\{x_i^*\\}[\/latex]).<\/p>\r\n<p id=\"fs-id1170571547592\">We look at some examples shortly. But, before we do, let\u2019s take a moment and talk about some specific choices for [latex]\\{x_i^*\\}[\/latex]. Although any choice for [latex]\\{x_i^*\\}[\/latex] gives us an estimate of the area under the curve, we don\u2019t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or low, we can select our value for [latex]\\{x_i^*\\}[\/latex] to guarantee one result or the other.<\/p>\r\n<p id=\"fs-id1170571711342\">If we want an overestimate, for example, we can choose [latex]\\{x_i^*\\}[\/latex] such that for [latex]i=1,2,3,\\cdots,n, \\, f(x_i^*)\\ge f(x)[\/latex] for all [latex]x\\in [x_{i-1},x_i][\/latex]. In other words, we choose [latex]\\{x_i^*\\}[\/latex] so that for [latex]i=1,2,3,\\cdots,n, \\, f(x_i^*)[\/latex] is the maximum function value on the interval [latex][x_{i-1},x_i][\/latex]. If we select [latex]\\{x_i^*\\}[\/latex] in this way, then the Riemann sum [latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex] is called an <strong>upper sum<\/strong>. Similarly, if we want an underestimate, we can choose [latex]\\{x_i^*\\}[\/latex] so that for [latex]i=1,2,3,\\cdots,n, \\, f(x_i^*)[\/latex] is the minimum function value on the interval [latex][x_{i-1},x_i][\/latex]. In this case, the associated Riemann sum is called a<strong> lower sum<\/strong>. Note that if [latex]f(x)[\/latex] is either increasing or decreasing throughout the interval [latex][a,b][\/latex], then the maximum and minimum values of the function occur at the endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.<\/p>\r\nTip: If a function is increasing over a closed interval, the right endpoints will be used to calculate the upper sum and the left endpoints the lower sum. If a function is decreasing over a closed interval, the right endpoints will be used for the lower sum and the left endpoints the upper sum.\r\n<div id=\"fs-id1170572274802\" class=\"textbook exercises\">\r\n<h3>example: Finding Lower and Upper Sums<\/h3>\r\nFind a lower sum for [latex]f(x)=10-x^2[\/latex] on [latex][1,2][\/latex]; use [latex]n=4[\/latex] subintervals.\r\n<div id=\"fs-id1170572274805\" class=\"exercise\">\r\n<div class=\"solution\">\r\n\r\n[reveal-answer q=\"448179\"]Show Solution[\/reveal-answer]\r\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],\u00a0(Figure 14) shows that a lower sum is obtained by using the right endpoints.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203921\/CNX_Calc_Figure_05_01_014.jpg\" alt=\"The graph of f(x) = 10 \u2212 x^2 from 0 to 2. It is set up for a right-end approximation of the area bounded by the curve and the x-axis on [1, 2], labeled a=x0 to x4. It shows a lower sum.\" width=\"487\" height=\"275\" \/> 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]\r\n<p id=\"fs-id1170572307267\">The Riemann sum is<\/p>\r\n\r\n<div id=\"fs-id1170572233863\" class=\"equation unnumbered\">[latex]\\begin{array}{ll}\\displaystyle\\sum_{k=1}^{4} (10-x^2)(0.25)&amp; =0.25[10-(1.25)^2+10-(1.5)^2+10-(1.75)^2+10-(2)^2] \\\\ &amp; =0.25[8.4375+7.75+6.9375+6] \\\\ &amp; =7.28 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572376172\">The area of 7.28 is a lower sum and an underestimate.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding Lower and Upper Sums.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qx-gvr8k8SY?controls=0&amp;start=825&amp;end=1020&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.1ApproximatingAreas825to1020_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1 Approximating Areas\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572376178\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<ol id=\"fs-id1170572376186\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find an upper sum for [latex]f(x)=10-x^2[\/latex] on [latex][1,2][\/latex]; let [latex]n=4[\/latex].<\/li>\r\n \t<li>Sketch the approximation.<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1170572376182\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"568309\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"568309\"]\r\n<p id=\"fs-id1170571653916\">[latex]f(x)[\/latex] is decreasing on [latex][1,2][\/latex], so the maximum function values occur at the left endpoints of the subintervals.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571653953\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571653953\"]\r\n<ol id=\"fs-id1170571653953\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Upper sum [latex]=8.0313[\/latex].<\/li>\r\n \t<li>[caption id=\"\" align=\"alignnone\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203924\/CNX_Calc_Figure_05_01_015.jpg\" alt=\"A graph of the function f(x) = 10 \u2212 x^2 from 0 to 2. It is set up for a right endpoint approximation over the area [1,2], which is labeled a=x0 to x4. It is an upper sum.\" width=\"487\" height=\"275\" \/> Figure 15.[\/caption]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572551906\" class=\"textbook exercises\">\r\n<h3>Example: Finding Lower and Upper Sums for [latex]f(x)= \\sin x[\/latex]<\/h3>\r\nFind a lower sum for [latex]f(x)= \\sin x[\/latex] over the interval [latex][a,b]=[0,\\frac{\\pi }{2}][\/latex]; let [latex]n=6[\/latex].\r\n[reveal-answer q=\"fs-id1170572626586\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572626586\"]\r\nLet\u2019s first look at the graph in Figure 16 to get a better idea of the area of interest.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203926\/CNX_Calc_Figure_05_01_016.jpg\" alt=\"A graph of the function y = sin(x) from 0 to pi. It is set up for a left endpoint approximation from 0 to pi\/2 and n=6. It is a lower sum.\" width=\"731\" height=\"238\" \/> Figure 16. The graph of [latex]y= \\sin x[\/latex] is divided into six regions: [latex]\\Delta x=\\frac{\\pi\/2}{6}=\\frac{\\pi}{12}[\/latex].[\/caption]\r\n<p id=\"fs-id1170571699095\">The intervals are [latex][0,\\frac{\\pi}{12}], \\, [\\frac{\\pi}{12},\\frac{\\pi}{6}], \\, [\\frac{\\pi}{6},\\frac{\\pi}{4}], \\, [\\frac{\\pi}{4},\\frac{\\pi}{3}], \\, [\\frac{\\pi}{3},\\frac{5\\pi}{12}][\/latex], and [latex][\\frac{5\\pi}{12},\\frac{\\pi}{2}][\/latex]. Note that [latex]f(x)= \\sin x[\/latex] is increasing on the interval [latex][0,\\frac{\\pi}{2}][\/latex], so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum [latex]\\displaystyle\\sum_{i=0}^{5} \\sin x_i(\\frac{\\pi}{12})[\/latex]. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}A &amp; \\approx \\sin (0)(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{12})(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{6})(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{4})(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{3})(\\frac{\\pi }{12})+ \\sin (\\frac{5\\pi }{12})(\\frac{\\pi }{12}) \\\\ &amp; =0.863 \\end{array}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding Lower and Upper Sums for [latex]f(x)= \\sin x[\/latex].\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qx-gvr8k8SY?controls=0&amp;start=1025&amp;end=1248&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.1ApproximatingAreas1025to1248_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1 Approximating Areas\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572554389\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the function [latex]f(x)= \\sin x[\/latex] over the interval [latex][0,\\frac{\\pi}{2}][\/latex], find an upper sum; let [latex]n=6[\/latex].\r\n\r\n[reveal-answer q=\"625320\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"625320\"]Follow the steps from the previous example.[\/hidden-answer]\r\n<div id=\"fs-id1170572554393\" class=\"exercise\">[reveal-answer q=\"fs-id1170571769558\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571769558\"]\r\n<p id=\"fs-id1170571769558\">[latex]A \\approx 1.125[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]210547[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use Riemann sums to approximate area<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1170572551818\" class=\"bc-section section\">\n<p id=\"fs-id1170572551824\">So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval [latex][x_{i-1},x_i][\/latex]. In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point [latex]x_i^*[\/latex]\u00a0in 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<div id=\"fs-id1170572376449\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A \\approx \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572376498\">A sum of this form is called a<strong> Riemann sum<\/strong>, named for the 19th-century mathematician Bernhard Riemann, who developed the idea.<\/p>\n<div id=\"fs-id1170572376504\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1170572376508\">Let [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]. A <strong>Riemann sum<\/strong> is defined for [latex]f(x)[\/latex] as<\/p>\n<div id=\"fs-id1170572444290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1170572344248\">Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as [latex]n[\/latex] get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of [latex]n[\/latex]. We are now ready to define the area under a curve in terms of Riemann sums.<\/p>\n<div id=\"fs-id1170572344262\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1170572344265\">Let [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]. Then, the <strong>area under the curve<\/strong> [latex]y=f(x)[\/latex] on [latex][a,b][\/latex] is given by<\/p>\n<div id=\"fs-id1170572229819\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1170572217482\" class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p><a href=\"http:\/\/archives.math.utk.edu\/visual.calculus\/4\/riemann_sums.4\/\" target=\"_blank\" rel=\"noopener\">See a graphical demonstration of the construction of a Riemann sum.<\/a><\/p>\n<\/div>\n<p id=\"fs-id1170572217494\">Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit of a function [latex]f(x)[\/latex] as [latex]x[\/latex] goes to infinity. Limits of sums are discussed in detail in the chapter on <a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/HTmjSAcf@2.46:7JGyeadv@3\/Introduction\">Sequences and Series<\/a>\u00a0in the second volume of this text; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums.<\/p>\n<p id=\"fs-id1170572217522\">Second, we must consider what to do if the expression converges to different limits for different choices of [latex]\\{x_i^*\\}[\/latex]. Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if [latex]f(x)[\/latex] is continuous on the closed interval [latex][a,b][\/latex], then [latex]\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex] exists and is unique (in other words, it does not depend on the choice of [latex]\\{x_i^*\\}[\/latex]).<\/p>\n<p id=\"fs-id1170571547592\">We look at some examples shortly. But, before we do, let\u2019s take a moment and talk about some specific choices for [latex]\\{x_i^*\\}[\/latex]. Although any choice for [latex]\\{x_i^*\\}[\/latex] gives us an estimate of the area under the curve, we don\u2019t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or low, we can select our value for [latex]\\{x_i^*\\}[\/latex] to guarantee one result or the other.<\/p>\n<p id=\"fs-id1170571711342\">If we want an overestimate, for example, we can choose [latex]\\{x_i^*\\}[\/latex] such that for [latex]i=1,2,3,\\cdots,n, \\, f(x_i^*)\\ge f(x)[\/latex] for all [latex]x\\in [x_{i-1},x_i][\/latex]. In other words, we choose [latex]\\{x_i^*\\}[\/latex] so that for [latex]i=1,2,3,\\cdots,n, \\, f(x_i^*)[\/latex] is the maximum function value on the interval [latex][x_{i-1},x_i][\/latex]. If we select [latex]\\{x_i^*\\}[\/latex] in this way, then the Riemann sum [latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex] is called an <strong>upper sum<\/strong>. Similarly, if we want an underestimate, we can choose [latex]\\{x_i^*\\}[\/latex] so that for [latex]i=1,2,3,\\cdots,n, \\, f(x_i^*)[\/latex] is the minimum function value on the interval [latex][x_{i-1},x_i][\/latex]. In this case, the associated Riemann sum is called a<strong> lower sum<\/strong>. Note that if [latex]f(x)[\/latex] is either increasing or decreasing throughout the interval [latex][a,b][\/latex], then the maximum and minimum values of the function occur at the endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.<\/p>\n<p>Tip: If a function is increasing over a closed interval, the right endpoints will be used to calculate the upper sum and the left endpoints the lower sum. If a function is decreasing over a closed interval, the right endpoints will be used for the lower sum and the left endpoints the upper sum.<\/p>\n<div id=\"fs-id1170572274802\" class=\"textbook exercises\">\n<h3>example: Finding Lower and Upper Sums<\/h3>\n<p>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<div id=\"fs-id1170572274805\" class=\"exercise\">\n<div class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q448179\">Show Solution<\/span><\/p>\n<div id=\"q448179\" class=\"hidden-answer\" style=\"display: none\">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],\u00a0(Figure 14) shows that a lower sum is obtained by using the right endpoints.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203921\/CNX_Calc_Figure_05_01_014.jpg\" alt=\"The graph of f(x) = 10 \u2212 x^2 from 0 to 2. It is set up for a right-end approximation of the area bounded by the curve and the x-axis on &#091;1, 2&#093;, labeled a=x0 to x4. It shows a lower sum.\" width=\"487\" height=\"275\" \/><\/p>\n<p class=\"wp-caption-text\">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.<\/p>\n<\/div>\n<p id=\"fs-id1170572307267\">The Riemann sum is<\/p>\n<div id=\"fs-id1170572233863\" class=\"equation unnumbered\">[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>&nbsp;<\/p>\n<p id=\"fs-id1170572376172\">The area of 7.28 is a lower sum and an underestimate.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding Lower and Upper Sums.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qx-gvr8k8SY?controls=0&amp;start=825&amp;end=1020&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.1ApproximatingAreas825to1020_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1 Approximating Areas&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572376178\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<ol id=\"fs-id1170572376186\" style=\"list-style-type: lower-alpha;\">\n<li>Find an upper sum for [latex]f(x)=10-x^2[\/latex] on [latex][1,2][\/latex]; let [latex]n=4[\/latex].<\/li>\n<li>Sketch the approximation.<\/li>\n<\/ol>\n<div id=\"fs-id1170572376182\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q568309\">Hint<\/span><\/p>\n<div id=\"q568309\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571653916\">[latex]f(x)[\/latex] is decreasing on [latex][1,2][\/latex], so the maximum function values occur at the left endpoints of the subintervals.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571653953\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571653953\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170571653953\" style=\"list-style-type: lower-alpha;\">\n<li>Upper sum [latex]=8.0313[\/latex].<\/li>\n<li>\n<div style=\"width: 497px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203924\/CNX_Calc_Figure_05_01_015.jpg\" alt=\"A graph of the function f(x) = 10 \u2212 x^2 from 0 to 2. It is set up for a right endpoint approximation over the area &#091;1,2&#093;, which is labeled a=x0 to x4. It is an upper sum.\" width=\"487\" height=\"275\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 15.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572551906\" class=\"textbook exercises\">\n<h3>Example: Finding Lower and Upper Sums for [latex]f(x)= \\sin x[\/latex]<\/h3>\n<p>Find a lower sum for [latex]f(x)= \\sin x[\/latex] over the interval [latex][a,b]=[0,\\frac{\\pi }{2}][\/latex]; let [latex]n=6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572626586\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572626586\" class=\"hidden-answer\" style=\"display: none\">\nLet\u2019s first look at the graph in Figure 16 to get a better idea of the area of interest.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203926\/CNX_Calc_Figure_05_01_016.jpg\" alt=\"A graph of the function y = sin(x) from 0 to pi. It is set up for a left endpoint approximation from 0 to pi\/2 and n=6. It is a lower sum.\" width=\"731\" height=\"238\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 16. The graph of [latex]y= \\sin x[\/latex] is divided into six regions: [latex]\\Delta x=\\frac{\\pi\/2}{6}=\\frac{\\pi}{12}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170571699095\">The intervals are [latex][0,\\frac{\\pi}{12}], \\, [\\frac{\\pi}{12},\\frac{\\pi}{6}], \\, [\\frac{\\pi}{6},\\frac{\\pi}{4}], \\, [\\frac{\\pi}{4},\\frac{\\pi}{3}], \\, [\\frac{\\pi}{3},\\frac{5\\pi}{12}][\/latex], and [latex][\\frac{5\\pi}{12},\\frac{\\pi}{2}][\/latex]. Note that [latex]f(x)= \\sin x[\/latex] is increasing on the interval [latex][0,\\frac{\\pi}{2}][\/latex], so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum [latex]\\displaystyle\\sum_{i=0}^{5} \\sin x_i(\\frac{\\pi}{12})[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}A & \\approx \\sin (0)(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{12})(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{6})(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{4})(\\frac{\\pi }{12})+ \\sin (\\frac{\\pi }{3})(\\frac{\\pi }{12})+ \\sin (\\frac{5\\pi }{12})(\\frac{\\pi }{12}) \\\\ & =0.863 \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding Lower and Upper Sums for [latex]f(x)= \\sin x[\/latex].<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qx-gvr8k8SY?controls=0&amp;start=1025&amp;end=1248&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.1ApproximatingAreas1025to1248_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1 Approximating Areas&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572554389\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the function [latex]f(x)= \\sin x[\/latex] over the interval [latex][0,\\frac{\\pi}{2}][\/latex], find an upper sum; let [latex]n=6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q625320\">Hint<\/span><\/p>\n<div id=\"q625320\" class=\"hidden-answer\" style=\"display: none\">Follow the steps from the previous example.<\/div>\n<\/div>\n<div id=\"fs-id1170572554393\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571769558\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571769558\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571769558\">[latex]A \\approx 1.125[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm210547\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=210547&theme=oea&iframe_resize_id=ohm210547&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1118\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>5.1 Approximating Areas. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.1 Approximating Areas\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1118","chapter","type-chapter","status-publish","hentry"],"part":1113,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1118","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1118\/revisions"}],"predecessor-version":[{"id":2636,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1118\/revisions\/2636"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1113"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1118\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1118"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1118"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1118"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1118"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}