{"id":427,"date":"2025-02-13T19:44:50","date_gmt":"2025-02-13T19:44:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/areas-between-curves-learn-it-2\/"},"modified":"2025-02-13T19:44:50","modified_gmt":"2025-02-13T19:44:50","slug":"areas-between-curves-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/areas-between-curves-learn-it-2\/","title":{"raw":"Areas Between Curves: Learn It 2","rendered":"Areas Between Curves: Learn It 2"},"content":{"raw":"\n<h2>Areas of Compound Regions<\/h2>\n<p id=\"fs-id1167793400867\">So far, we have required [latex]f(x)\\ge g(x)[\/latex] over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>finding the area of a region between curves that cross<\/h3>\n<p id=\"fs-id1167793589368\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions over an interval [latex]\\left[a,b\\right].[\/latex] Let [latex]R[\/latex] denote the region between the graphs of [latex]f(x)[\/latex] and [latex]g(x),[\/latex] and be bounded on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Then, the area of [latex]R[\/latex] is given by:<\/p>\n<div id=\"fs-id1167794162812\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1167793452414\">In practice, applying this theorem requires us to break up the interval [latex]\\left[a,b\\right][\/latex] and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.<\/p>\n<section class=\"textbox example\">\n<p>If [latex]R[\/latex] is the region between the graphs of the functions [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] over the interval [latex]\\left[0,\\pi \\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793510078\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793510078\"]\n\n<p id=\"fs-id1167793510078\">The region is depicted in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"348\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212620\/CNX_Calc_Figure_06_01_005.jpg\" alt=\"This figure is has two graphs. They are the functions f(x) = sinx and g(x)= cosx. They are both periodic functions that resemble waves. There are two shaded areas between the graphs. The first shaded area is labeled \u201cR1\u201d and has g(x) above f(x). This region begins at the y-axis and stops where the curves intersect. The second region is labeled \u201cR2\u201d and begins at the intersection with f(x) above g(x). The shaded region stops at x=pi.\" width=\"348\" height=\"347\"> Figure 5. The region between two curves can be broken into two sub-regions.[\/caption]\n\n<p id=\"fs-id1167793420158\">The graphs of the functions intersect at [latex]x=\\pi \\text{\/}4.[\/latex]<\/p>\n<p>For [latex]x\\in \\left[0,\\pi \\text{\/}4\\right],[\/latex] [latex] \\cos x\\ge \\sin x,[\/latex] so:<\/p>\n<div id=\"fs-id1167794325185\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\cos x- \\sin x[\/latex]<\/div>\n<p id=\"fs-id1167793258792\">On the other hand, for [latex]x\\in \\left[\\pi \\text{\/}4,\\pi \\right],[\/latex] [latex] \\sin x\\ge \\cos x,[\/latex] so:<\/p>\n<div id=\"fs-id1167794026884\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\sin x- \\cos x[\/latex]<\/div>\n<p id=\"fs-id1167794040499\">Then:<\/p>\n<div id=\"fs-id1167794040502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A&amp; ={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }| \\sin x- \\cos x|dx={\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}( \\cos x- \\sin x)dx+{\\displaystyle\\int }_{\\pi \\text{\/}4}^{\\pi }( \\sin x- \\cos x)dx\\hfill \\\\ &amp; ={\\left[ \\sin x+ \\cos x\\right]|}_{0}^{\\pi \\text{\/}4}+{\\left[\\text{\u2212} \\cos x- \\sin x\\right]|}_{\\pi \\text{\/}4}^{\\pi }\\hfill \\\\ &amp; =(\\sqrt{2}-1)+(1+\\sqrt{2})=2\\sqrt{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794117704\">The area of the region is [latex]2\\sqrt{2}[\/latex] units<sup>2<\/sup>.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793581456\">Consider the region depicted in the following figure. Find the area of [latex]R.[\/latex]<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212625\/CNX_Calc_Figure_06_01_007.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = squareroot of x and g(x)= 3\/2 \u2013 x\/2. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The shaded area is between x=0 and x=3.\" width=\"250\" height=\"235\"> Figure 7.[\/caption]\n\n<p><br>\n[reveal-answer q=\"967767\"]Hint[\/reveal-answer]<br>\n[hidden-answer a=\"967767\"]The two curves intersect at [latex]x=1.[\/latex][\/hidden-answer]<\/p>\n<p>[reveal-answer q=\"fs-id1167793262779\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793262779\"]<\/p>\n<p id=\"fs-id1167793262779\">[latex]\\frac{5}{3}[\/latex] units<sup>2<\/sup><\/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\/cm88bTFvRU4?controls=0&amp;start=1269&amp;end=1353&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>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\/2.1AreaBetweenCurves1269to1353_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.1 Area Between Curves\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<h2>Areas of Compound Regions<\/h2>\n<p id=\"fs-id1167793400867\">So far, we have required [latex]f(x)\\ge g(x)[\/latex] over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>finding the area of a region between curves that cross<\/h3>\n<p id=\"fs-id1167793589368\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions over an interval [latex]\\left[a,b\\right].[\/latex] Let [latex]R[\/latex] denote the region between the graphs of [latex]f(x)[\/latex] and [latex]g(x),[\/latex] and be bounded on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Then, the area of [latex]R[\/latex] is given by:<\/p>\n<div id=\"fs-id1167794162812\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1167793452414\">In practice, applying this theorem requires us to break up the interval [latex]\\left[a,b\\right][\/latex] and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.<\/p>\n<section class=\"textbox example\">\n<p>If [latex]R[\/latex] is the region between the graphs of the functions [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] over the interval [latex]\\left[0,\\pi \\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793510078\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793510078\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793510078\">The region is depicted in the following figure.<\/p>\n<figure style=\"width: 348px\" 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\/11212620\/CNX_Calc_Figure_06_01_005.jpg\" alt=\"This figure is has two graphs. They are the functions f(x) = sinx and g(x)= cosx. They are both periodic functions that resemble waves. There are two shaded areas between the graphs. The first shaded area is labeled \u201cR1\u201d and has g(x) above f(x). This region begins at the y-axis and stops where the curves intersect. The second region is labeled \u201cR2\u201d and begins at the intersection with f(x) above g(x). The shaded region stops at x=pi.\" width=\"348\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 5. The region between two curves can be broken into two sub-regions.<\/figcaption><\/figure>\n<p id=\"fs-id1167793420158\">The graphs of the functions intersect at [latex]x=\\pi \\text{\/}4.[\/latex]<\/p>\n<p>For [latex]x\\in \\left[0,\\pi \\text{\/}4\\right],[\/latex] [latex]\\cos x\\ge \\sin x,[\/latex] so:<\/p>\n<div id=\"fs-id1167794325185\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\cos x- \\sin x[\/latex]<\/div>\n<p id=\"fs-id1167793258792\">On the other hand, for [latex]x\\in \\left[\\pi \\text{\/}4,\\pi \\right],[\/latex] [latex]\\sin x\\ge \\cos x,[\/latex] so:<\/p>\n<div id=\"fs-id1167794026884\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\sin x- \\cos x[\/latex]<\/div>\n<p id=\"fs-id1167794040499\">Then:<\/p>\n<div id=\"fs-id1167794040502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A& ={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }| \\sin x- \\cos x|dx={\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}( \\cos x- \\sin x)dx+{\\displaystyle\\int }_{\\pi \\text{\/}4}^{\\pi }( \\sin x- \\cos x)dx\\hfill \\\\ & ={\\left[ \\sin x+ \\cos x\\right]|}_{0}^{\\pi \\text{\/}4}+{\\left[\\text{\u2212} \\cos x- \\sin x\\right]|}_{\\pi \\text{\/}4}^{\\pi }\\hfill \\\\ & =(\\sqrt{2}-1)+(1+\\sqrt{2})=2\\sqrt{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794117704\">The area of the region is [latex]2\\sqrt{2}[\/latex] units<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793581456\">Consider the region depicted in the following figure. Find the area of [latex]R.[\/latex]<\/p>\n<figure style=\"width: 250px\" 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\/11212625\/CNX_Calc_Figure_06_01_007.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = squareroot of x and g(x)= 3\/2 \u2013 x\/2. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The shaded area is between x=0 and x=3.\" width=\"250\" height=\"235\" \/><figcaption class=\"wp-caption-text\">Figure 7.<\/figcaption><\/figure>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q967767\">Hint<\/button><\/p>\n<div id=\"q967767\" class=\"hidden-answer\" style=\"display: none\">The two curves intersect at [latex]x=1.[\/latex]<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793262779\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793262779\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793262779\">[latex]\\frac{5}{3}[\/latex] units<sup>2<\/sup><\/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\/cm88bTFvRU4?controls=0&amp;start=1269&amp;end=1353&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/2.1AreaBetweenCurves1269to1353_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.1 Area Between Curves&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":421,"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\/427"}],"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\/427\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/421"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/427\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=427"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=427"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=427"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=427"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}