{"id":1124,"date":"2021-06-30T17:01:56","date_gmt":"2021-06-30T17:01:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/average-value-of-a-function\/"},"modified":"2022-03-19T03:17:06","modified_gmt":"2022-03-19T03:17:06","slug":"average-value-of-a-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/average-value-of-a-function\/","title":{"raw":"Average Value of a Function","rendered":"Average Value of a Function"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the average value of a function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1170571653448\" class=\"bc-section section\">\r\n<p id=\"fs-id1170571653453\">We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the following test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you want to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores. In this case, there are six test scores. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170571653460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{89+90+56+78+100+69}{6}=\\dfrac{482}{6}\\approx 80.33[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571788116\">Therefore, your average test grade is approximately 80.33, which translates to a B\u2212 at most schools.<\/p>\r\n<p id=\"fs-id1170571788122\">Suppose, however, that we have a function [latex]v(t)[\/latex] that gives us the speed of an object at any time [latex]t[\/latex], and we want to find the object\u2019s average speed. The function [latex]v(t)[\/latex] takes on an infinite number of values, so we can\u2019t use the process just described. Fortunately, we can use a definite integral to find the <strong>average value of a function<\/strong> such as this.<\/p>\r\n<p id=\"fs-id1170571597403\">Let [latex]f(x)[\/latex] be continuous over the interval [latex][a,b][\/latex] and let [latex][a,b][\/latex] be divided into [latex]n[\/latex] subintervals of width [latex]\\Delta x=\\frac{(b-a)}{n}[\/latex]. Choose a representative [latex]x_i^*[\/latex] in each subinterval and calculate [latex]f(x_i^*)[\/latex] for [latex]i=1,2, \\cdots , n[\/latex]. In other words, consider each [latex]f(x_i^*)[\/latex] as a sampling of the function over each subinterval. The average value of the function may then be approximated as<\/p>\r\n\r\n<div id=\"fs-id1170572480573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{n}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572369399\">which is basically the same expression used to calculate the average of discrete values.<\/p>\r\n<p id=\"fs-id1170572369402\">But we know [latex]\\Delta x=\\frac{b-a}{n}[\/latex], so [latex]n=\\frac{b-a}{\\Delta x}[\/latex], and we get<\/p>\r\n\r\n<div id=\"fs-id1170572373417\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{n}=\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{\\dfrac{(b-a)}{\\Delta x}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nFollowing through with the algebra, the numerator is a sum that is represented as [latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)[\/latex], and we are dividing by a fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value of the function is given by\r\n<div id=\"fs-id1170571734014\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\frac{\\displaystyle\\sum_{i=1}^{n} f(x_i^*)}{\\dfrac{\\left(b-a\\right)}{\\Delta x}} &amp; =\\left(\\dfrac{\\Delta x}{b-a}\\right)\\displaystyle\\sum_{i=1}^{n} f(x_i^*) \\\\ &amp; =\\left(\\dfrac{1}{b-a}\\right)\\displaystyle\\sum_{i=1}^{n} f(x_i^*) \\Delta x \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572274944\">This is a Riemann sum. Then, to get the <em>exact<\/em> average value, take the limit as [latex]n[\/latex] goes to infinity. Thus, the average value of a function is given by<\/p>\r\n\r\n<div id=\"fs-id1170572274957\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{1}{b-a}\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i) \\Delta x=\\dfrac{1}{b-a} \\displaystyle\\int_a^b f(x) dx[\/latex].<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1170572510086\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572510089\">Let [latex]f(x)[\/latex] be continuous over the interval [latex][a,b][\/latex]. Then, the <strong>average value of the function<\/strong> [latex]f(x)[\/latex] (or [latex]f_{\\text{ave}}[\/latex]) on [latex][a,b][\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1170571712537\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f_{\\text{ave}}=\\dfrac{1}{b-a} \\displaystyle\\int_a^b f(x) dx[\/latex].<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571712593\" class=\"textbox exercises\">\r\n<h3>Example: Finding the Average Value of a Linear Function<\/h3>\r\n<p id=\"fs-id1170571712602\">Find the average value of [latex]f(x)=x+1[\/latex] over the interval [latex][0,5][\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572443612\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572443612\"]\r\n<p id=\"fs-id1170572443612\">First, graph the function on the stated interval, as shown in Figure 11.<\/p>\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\/11204037\/CNX_Calc_Figure_05_02_017.jpg\" alt=\"A graph in quadrant one showing the shaded area under the function f(x) = x + 1 over [0,5].\" width=\"487\" height=\"284\" \/> Figure 11. The graph shows the area under the function [latex]f(x)=x+1[\/latex] over [latex][0,5][\/latex].[\/caption]\r\n<p id=\"fs-id1170572419232\">The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid [latex]A=\\frac{1}{2}h(a+b)[\/latex], where [latex]h[\/latex] represents height, and [latex]a[\/latex] and [latex]b[\/latex] represent the two parallel sides. Then,<\/p>\r\n\r\n<div id=\"fs-id1170572419282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^5 x+1 dx &amp; =\\frac{1}{2}h(a+b) \\\\ &amp; =\\frac{1}{2} \\cdot 5 \\cdot (1+6) \\\\ &amp; =\\frac{35}{2} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571699476\">Thus the average value of the function is<\/p>\r\n\r\n<div id=\"fs-id1170571699479\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{5-0} \\displaystyle\\int_0^5 x+1 dx = \\frac{1}{5} \\cdot \\frac{35}{2}=\\frac{7}{2}.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding the Average Value of a Linear Function.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tto0E7yOSLo?controls=0&amp;start=1427&amp;end=1559&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\/5.2TheDefiniteIntegral1427to1559_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.2 The Definite Integral\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170571678845\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the average value of [latex]f(x)=6-2x[\/latex] over the interval [latex][0,3][\/latex].\r\n\r\n[reveal-answer q=\"299231\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"299231\"]\r\n\r\nUse the average value formula, and use geometry to evaluate the integral.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"7086644\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7086644\"]\r\n<p id=\"fs-id1170572601248\">3<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20448[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the average value of a function<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1170571653448\" class=\"bc-section section\">\n<p id=\"fs-id1170571653453\">We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the following test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you want to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores. In this case, there are six test scores. Thus,<\/p>\n<div id=\"fs-id1170571653460\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{89+90+56+78+100+69}{6}=\\dfrac{482}{6}\\approx 80.33[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571788116\">Therefore, your average test grade is approximately 80.33, which translates to a B\u2212 at most schools.<\/p>\n<p id=\"fs-id1170571788122\">Suppose, however, that we have a function [latex]v(t)[\/latex] that gives us the speed of an object at any time [latex]t[\/latex], and we want to find the object\u2019s average speed. The function [latex]v(t)[\/latex] takes on an infinite number of values, so we can\u2019t use the process just described. Fortunately, we can use a definite integral to find the <strong>average value of a function<\/strong> such as this.<\/p>\n<p id=\"fs-id1170571597403\">Let [latex]f(x)[\/latex] be continuous over the interval [latex][a,b][\/latex] and let [latex][a,b][\/latex] be divided into [latex]n[\/latex] subintervals of width [latex]\\Delta x=\\frac{(b-a)}{n}[\/latex]. Choose a representative [latex]x_i^*[\/latex] in each subinterval and calculate [latex]f(x_i^*)[\/latex] for [latex]i=1,2, \\cdots , n[\/latex]. In other words, consider each [latex]f(x_i^*)[\/latex] as a sampling of the function over each subinterval. The average value of the function may then be approximated as<\/p>\n<div id=\"fs-id1170572480573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{n}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572369399\">which is basically the same expression used to calculate the average of discrete values.<\/p>\n<p id=\"fs-id1170572369402\">But we know [latex]\\Delta x=\\frac{b-a}{n}[\/latex], so [latex]n=\\frac{b-a}{\\Delta x}[\/latex], and we get<\/p>\n<div id=\"fs-id1170572373417\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{n}=\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{\\dfrac{(b-a)}{\\Delta x}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Following through with the algebra, the numerator is a sum that is represented as [latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)[\/latex], and we are dividing by a fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value of the function is given by<\/p>\n<div id=\"fs-id1170571734014\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\frac{\\displaystyle\\sum_{i=1}^{n} f(x_i^*)}{\\dfrac{\\left(b-a\\right)}{\\Delta x}} & =\\left(\\dfrac{\\Delta x}{b-a}\\right)\\displaystyle\\sum_{i=1}^{n} f(x_i^*) \\\\ & =\\left(\\dfrac{1}{b-a}\\right)\\displaystyle\\sum_{i=1}^{n} f(x_i^*) \\Delta x \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572274944\">This is a Riemann sum. Then, to get the <em>exact<\/em> average value, take the limit as [latex]n[\/latex] goes to infinity. Thus, the average value of a function is given by<\/p>\n<div id=\"fs-id1170572274957\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{1}{b-a}\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i) \\Delta x=\\dfrac{1}{b-a} \\displaystyle\\int_a^b f(x) dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1170572510086\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170572510089\">Let [latex]f(x)[\/latex] be continuous over the interval [latex][a,b][\/latex]. Then, the <strong>average value of the function<\/strong> [latex]f(x)[\/latex] (or [latex]f_{\\text{ave}}[\/latex]) on [latex][a,b][\/latex] is given by<\/p>\n<div id=\"fs-id1170571712537\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f_{\\text{ave}}=\\dfrac{1}{b-a} \\displaystyle\\int_a^b f(x) dx[\/latex].<\/div>\n<\/div>\n<div id=\"fs-id1170571712593\" class=\"textbox exercises\">\n<h3>Example: Finding the Average Value of a Linear Function<\/h3>\n<p id=\"fs-id1170571712602\">Find the average value of [latex]f(x)=x+1[\/latex] over the interval [latex][0,5][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572443612\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572443612\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572443612\">First, graph the function on the stated interval, as shown in Figure 11.<\/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\/11204037\/CNX_Calc_Figure_05_02_017.jpg\" alt=\"A graph in quadrant one showing the shaded area under the function f(x) = x + 1 over &#091;0,5&#093;.\" width=\"487\" height=\"284\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 11. The graph shows the area under the function [latex]f(x)=x+1[\/latex] over [latex][0,5][\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170572419232\">The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid [latex]A=\\frac{1}{2}h(a+b)[\/latex], where [latex]h[\/latex] represents height, and [latex]a[\/latex] and [latex]b[\/latex] represent the two parallel sides. Then,<\/p>\n<div id=\"fs-id1170572419282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^5 x+1 dx & =\\frac{1}{2}h(a+b) \\\\ & =\\frac{1}{2} \\cdot 5 \\cdot (1+6) \\\\ & =\\frac{35}{2} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571699476\">Thus the average value of the function is<\/p>\n<div id=\"fs-id1170571699479\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{5-0} \\displaystyle\\int_0^5 x+1 dx = \\frac{1}{5} \\cdot \\frac{35}{2}=\\frac{7}{2}.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Average Value of a Linear Function.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tto0E7yOSLo?controls=0&amp;start=1427&amp;end=1559&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\/5.2TheDefiniteIntegral1427to1559_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.2 The Definite Integral&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170571678845\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the average value of [latex]f(x)=6-2x[\/latex] over the interval [latex][0,3][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q299231\">Hint<\/span><\/p>\n<div id=\"q299231\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the average value formula, and use geometry to evaluate the integral.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7086644\">Show Solution<\/span><\/p>\n<div id=\"q7086644\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572601248\">3<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20448\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20448&theme=oea&iframe_resize_id=ohm20448&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-1124\">\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>5.2 The Definite Integral. <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":11,"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\":\"5.2 The Definite Integral\",\"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-1124","chapter","type-chapter","status-publish","hentry"],"part":1113,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1124","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":3,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1124\/revisions"}],"predecessor-version":[{"id":1646,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1124\/revisions\/1646"}],"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\/1124\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1124"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1124"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1124"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1124"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}