{"id":380,"date":"2025-02-13T19:44:28","date_gmt":"2025-02-13T19:44:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-areas-fresh-take\/"},"modified":"2025-02-13T19:44:28","modified_gmt":"2025-02-13T19:44:28","slug":"approximating-areas-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-areas-fresh-take\/","title":{"raw":"Approximating Areas: Fresh Take","rendered":"Approximating Areas: Fresh Take"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Estimate the area under a curve by adding up the areas of rectangles<\/li>\n\t<li>Estimate the area under a curve using Riemann sums<\/li>\n<\/ul>\n<\/section>\n<h2>Approximating Area<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Area under a curve can be approximated using rectangles<\/li>\n\t<li class=\"whitespace-normal break-words\">Two main methods: left-endpoint and right-endpoint approximations<\/li>\n\t<li class=\"whitespace-normal break-words\">Increasing the number of rectangles improves accuracy<\/li>\n\t<li class=\"whitespace-normal break-words\">Riemann sums formalize this approximation process<\/li>\n\t<li class=\"whitespace-normal break-words\">Partitions:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Divide interval [latex][a,b][\/latex] into n subintervals<\/li>\n\t<li class=\"whitespace-normal break-words\">Regular partition: subintervals of equal width [latex]\\Delta x = \\frac{(b-a)}{n}[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Left-Endpoint Approximation:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]L_n = \\sum_{i=1}^n f(x_{i-1})\\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Rectangle heights based on function value at left endpoint<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Right-Endpoint Approximation:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]R_n = \\sum_{i=1}^n f(x_i)\\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Rectangle heights based on function value at right endpoint<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Convergence:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">As [latex]n[\/latex] increases, approximations generally become more accurate<\/li>\n\t<li class=\"whitespace-normal break-words\">Left and right approximations often converge to the true area<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">Write in sigma notation and evaluate the sum of terms [latex]2^i[\/latex]&nbsp;for [latex]i=3,4,5,6[\/latex]. [reveal-answer q=\"fs-id1170572133846\"]Show Solution[\/reveal-answer] [hidden-answer a=\"fs-id1170572133846\"]\n\n<p id=\"fs-id1170572133846\">[latex]\\displaystyle\\sum_{i=3}^{6} 2^i=2^3+2^4+2^5+2^6=120[\/latex]<\/p>\n<p id=\"fs-id1170572167964\">[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the sum of the values of [latex]f(x)=x^3[\/latex] over the integers [latex]1,2,3,\\cdots,10[\/latex].<\/p>\n\n[reveal-answer q=\"fs-id1170571599435\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571599435\"]\n\n<p id=\"fs-id1170571599435\">Using the formula, we have<\/p>\n<div id=\"fs-id1170572224759\" class=\"equation unnumbered\">[latex]\\begin{array}{ll}\\displaystyle\\sum_{i=0}^{10} i^3 &amp; =\\frac{(10)^2(10+1)^2}{4} \\\\ &amp; =\\frac{100(121)}{4} \\\\ &amp; =3025 \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find left and right-endpoint approximations for [latex]\\int_0^2 x^2 dx[\/latex] with [latex]n = 4[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><br>\n[reveal-answer q=\"666963\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"666963\"]<\/p>\n<p>Partition:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta x = \\frac{(2-0)}{4} = 0.5[\/latex]<\/p>\n<p>Subintervals:<\/p>\n<p>[latex][0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2][\/latex]<\/p>\n<p><br>\nLeft-Endpoint Approximation:<\/p>\n<p style=\"text-align: center;\">[latex]L_4 = (0^2)(0.5) + (0.5^2)(0.5) + (1^2)(0.5) + (1.5^2)(0.5)[\/latex]<br>\n[latex]= 0 + 0.125 + 0.5 + 1.125 = 1.75[\/latex]<\/p>\n<p>Right-Endpoint Approximation:<\/p>\n<p style=\"text-align: center;\">[latex]R_4 = (0.5^2)(0.5) + (1^2)(0.5) + (1.5^2)(0.5) + (2^2)(0.5)[\/latex]<br>\n[latex]= 0.125 + 0.5 + 1.125 + 2 = 3.75[\/latex]<\/p>\n<p>Comparison:<\/p>\n<p style=\"text-align: center;\">[latex]L_4 = 1.75 &lt; \\text{ True Area } &lt; R_4 = 3.75[\/latex]<\/p>\n<p>Note: The true value is [latex]\\int_0^2 x^2 dx = [\\frac{1}{3}x^3]_0^2 = \\frac{8}{3} \\approx 2.67[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\n<\/section>\n<h2>Riemann Sums<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Riemann sums generalize left and right endpoint approximations. They allow evaluation of the function at any point within each subinterval<\/li>\n\t<li class=\"whitespace-normal break-words\">As the number of subintervals increases, the approximation improves<\/li>\n\t<li class=\"whitespace-normal break-words\">Riemann sums can be used to define the area under a curve<\/li>\n\t<li class=\"whitespace-normal break-words\">Riemann Sum Definition:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex] where [latex]x_i^*[\/latex] is any point in the subinterval [latex][x_{i-1},x_i][\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Area Under a Curve:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]A=\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Upper and Lower Sums:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Upper sum:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Choose [latex]x_i^*[\/latex] for maximum [latex]f(x_i^*)[\/latex] in each subinterval<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Lower sum:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Choose [latex]x_i^*[\/latex] for minimum [latex]f(x_i^*)[\/latex] in each subinterval<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Convergence:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">For continuous functions, the limit of Riemann sums is unique<\/li>\n\t<li class=\"whitespace-normal break-words\">The limit doesn't depend on the choice of [latex]x_i^*[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\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].<br>\n[reveal-answer q=\"fs-id1170572626586\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572626586\"]<br>\nLet\u2019s first look at the graph in Figure 16 to get a better idea of the area of interest.<\/p>\n\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]\n\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 &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]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\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\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p style=\"text-align: center;\">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 \"5.1 Approximating Areas\" here (opens in new window)<\/a>. [\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\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\n[reveal-answer q=\"fs-id1170571769558\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571769558\"]\n\n<p id=\"fs-id1170571769558\">[latex]A \\approx 1.125[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Estimate the area under a curve by adding up the areas of rectangles<\/li>\n<li>Estimate the area under a curve using Riemann sums<\/li>\n<\/ul>\n<\/section>\n<h2>Approximating Area<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Area under a curve can be approximated using rectangles<\/li>\n<li class=\"whitespace-normal break-words\">Two main methods: left-endpoint and right-endpoint approximations<\/li>\n<li class=\"whitespace-normal break-words\">Increasing the number of rectangles improves accuracy<\/li>\n<li class=\"whitespace-normal break-words\">Riemann sums formalize this approximation process<\/li>\n<li class=\"whitespace-normal break-words\">Partitions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Divide interval [latex][a,b][\/latex] into n subintervals<\/li>\n<li class=\"whitespace-normal break-words\">Regular partition: subintervals of equal width [latex]\\Delta x = \\frac{(b-a)}{n}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Left-Endpoint Approximation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]L_n = \\sum_{i=1}^n f(x_{i-1})\\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Rectangle heights based on function value at left endpoint<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Right-Endpoint Approximation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]R_n = \\sum_{i=1}^n f(x_i)\\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Rectangle heights based on function value at right endpoint<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Convergence:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">As [latex]n[\/latex] increases, approximations generally become more accurate<\/li>\n<li class=\"whitespace-normal break-words\">Left and right approximations often converge to the true area<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">Write in sigma notation and evaluate the sum of terms [latex]2^i[\/latex]&nbsp;for [latex]i=3,4,5,6[\/latex]. <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572133846\">Show Solution<\/button> <\/p>\n<div id=\"qfs-id1170572133846\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572133846\">[latex]\\displaystyle\\sum_{i=3}^{6} 2^i=2^3+2^4+2^5+2^6=120[\/latex]<\/p>\n<p id=\"fs-id1170572167964\"><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the sum of the values of [latex]f(x)=x^3[\/latex] over the integers [latex]1,2,3,\\cdots,10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571599435\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571599435\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571599435\">Using the formula, we have<\/p>\n<div id=\"fs-id1170572224759\" class=\"equation unnumbered\">[latex]\\begin{array}{ll}\\displaystyle\\sum_{i=0}^{10} i^3 & =\\frac{(10)^2(10+1)^2}{4} \\\\ & =\\frac{100(121)}{4} \\\\ & =3025 \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find left and right-endpoint approximations for [latex]\\int_0^2 x^2 dx[\/latex] with [latex]n = 4[\/latex].<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q666963\">Show Answer<\/button><\/p>\n<div id=\"q666963\" class=\"hidden-answer\" style=\"display: none\">\n<p>Partition:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta x = \\frac{(2-0)}{4} = 0.5[\/latex]<\/p>\n<p>Subintervals:<\/p>\n<p>[latex][0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2][\/latex]<\/p>\n<p>\nLeft-Endpoint Approximation:<\/p>\n<p style=\"text-align: center;\">[latex]L_4 = (0^2)(0.5) + (0.5^2)(0.5) + (1^2)(0.5) + (1.5^2)(0.5)[\/latex]<br \/>\n[latex]= 0 + 0.125 + 0.5 + 1.125 = 1.75[\/latex]<\/p>\n<p>Right-Endpoint Approximation:<\/p>\n<p style=\"text-align: center;\">[latex]R_4 = (0.5^2)(0.5) + (1^2)(0.5) + (1.5^2)(0.5) + (2^2)(0.5)[\/latex]<br \/>\n[latex]= 0.125 + 0.5 + 1.125 + 2 = 3.75[\/latex]<\/p>\n<p>Comparison:<\/p>\n<p style=\"text-align: center;\">[latex]L_4 = 1.75 < \\text{ True Area } < R_4 = 3.75[\/latex]<\/p>\n<p>Note: The true value is [latex]\\int_0^2 x^2 dx = [\\frac{1}{3}x^3]_0^2 = \\frac{8}{3} \\approx 2.67[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<h2>Riemann Sums<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Riemann sums generalize left and right endpoint approximations. They allow evaluation of the function at any point within each subinterval<\/li>\n<li class=\"whitespace-normal break-words\">As the number of subintervals increases, the approximation improves<\/li>\n<li class=\"whitespace-normal break-words\">Riemann sums can be used to define the area under a curve<\/li>\n<li class=\"whitespace-normal break-words\">Riemann Sum Definition:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex] where [latex]x_i^*[\/latex] is any point in the subinterval [latex][x_{i-1},x_i][\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Area Under a Curve:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]A=\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Upper and Lower Sums:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Upper sum:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Choose [latex]x_i^*[\/latex] for maximum [latex]f(x_i^*)[\/latex] in each subinterval<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Lower sum:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Choose [latex]x_i^*[\/latex] for minimum [latex]f(x_i^*)[\/latex] in each subinterval<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Convergence:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For continuous functions, the limit of Riemann sums is unique<\/li>\n<li class=\"whitespace-normal break-words\">The limit doesn&#8217;t depend on the choice of [latex]x_i^*[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572626586\">Show Solution<\/button><\/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<figure style=\"width: 731px\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\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]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/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\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p style=\"text-align: center;\">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<\/section>\n<section class=\"textbox example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571769558\">Show Solution<\/button><\/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<\/section>\n","protected":false},"author":6,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":371,"module-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\/380"}],"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\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/380\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/371"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/380\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=380"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=380"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=380"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=380"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}