{"id":430,"date":"2025-02-13T19:44:51","date_gmt":"2025-02-13T19:44:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/areas-between-curves-fresh-take\/"},"modified":"2025-02-13T19:44:51","modified_gmt":"2025-02-13T19:44:51","slug":"areas-between-curves-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/areas-between-curves-fresh-take\/","title":{"raw":"Areas Between Curves: Fresh Take","rendered":"Areas Between Curves: Fresh Take"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Calculate the area between two curves by integrating with respect to [latex]x[\/latex]<\/li>\n\t<li>Calculate the area of a compound region<\/li>\n\t<li>Calculate the area between two curves by integrating with respect to [latex]y[\/latex]<\/li>\n\t<li>Determine the most effective variable, [latex]x[\/latex] or [latex]y[\/latex], for integration based on the curves\u2019 orientation<\/li>\n<\/ul>\n<\/section>\n<h2>Area of a Region between Two Curves<\/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\">Extends the concept of definite integrals to calculate areas of more complex regions<\/li>\n\t<li class=\"whitespace-normal break-words\">For continuous functions [latex]f(x)[\/latex] and [latex]g(x)[\/latex] where f(x) \u2265 g(x) on [latex][a,b][\/latex]:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]A = \\int_a^b [f(x) - g(x)] dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Area represents the difference between the upper and lower function, integrated over the interval<\/li>\n\t<li class=\"whitespace-normal break-words\">Process:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Identify the upper and lower functions<\/li>\n\t<li class=\"whitespace-normal break-words\">Determine the interval of integration<\/li>\n\t<li class=\"whitespace-normal break-words\">Set up and evaluate the definite integral<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Often necessary to solve [latex]f(x) = g(x)[\/latex] to determine integration limits<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>If [latex]R[\/latex] is the region bounded by the graphs of the functions [latex]f(x)=\\frac{x}{2}+5[\/latex] and [latex]g(x)=x+\\frac{1}{2}[\/latex] over the interval [latex]\\left[1,5\\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793933114\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793933114\"]\n\n<p id=\"fs-id1167793933114\">[latex]12[\/latex] units<sup>2<\/sup><\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>If [latex]R[\/latex] is the region bounded above by the graph of the function [latex]f(x)=x[\/latex] and below by the graph of the function [latex]g(x)={x}^{4},[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793276952\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793276952\"]\n\n<p id=\"fs-id1167793276952\">[latex]\\frac{3}{10}[\/latex] unit<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=616&amp;end=707&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\/2.1AreaBetweenCurves616to707_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<h2>Areas of Compound Regions<\/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\">Extends area calculations to regions where functions intersect<\/li>\n\t<li class=\"whitespace-normal break-words\">Key formula for regions bounded by intersecting curves [latex]f(x)[\/latex] and [latex]g(x)[\/latex] on [latex][a,b][\/latex]:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]A = \\int_a^b |f(x) - g(x)| dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Absolute value in the formula accounts for changes in which function is greater<\/li>\n\t<li class=\"whitespace-normal break-words\">Process:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Identify intersection points of the functions<\/li>\n\t<li class=\"whitespace-normal break-words\">Break the interval into subintervals where one function is consistently greater<\/li>\n\t<li class=\"whitespace-normal break-words\">Set up and evaluate separate integrals for each subinterval<\/li>\n\t<li class=\"whitespace-normal break-words\">Sum the results to get the total area<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\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[\\frac{\\pi}{2},2\\pi \\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<p><br>\n[reveal-answer q=\"118401\"]Hint[\/reveal-answer]<br>\n[hidden-answer a=\"118401\"]The two curves intersect at [latex]x=\\frac{(5\\pi )}{4}.[\/latex][\/hidden-answer]<\/p>\n\n[reveal-answer q=\"fs-id1167793932268\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793932268\"]\n\n<p id=\"fs-id1167793932268\">[latex]2+2\\sqrt{2}[\/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=1007&amp;end=1168&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\/2.1AreaBetweenCurves1007to1168_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<section class=\"textbox example\">\n<p id=\"fs-id1167793517455\">Consider the region depicted below. Find the area of [latex]R.[\/latex]<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"229\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212623\/CNX_Calc_Figure_06_01_006.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = x^2 and g(x)= 2-x. 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 region is labeled R. The shaded area is between x=0 and x=2.\" width=\"229\" height=\"242\"> Figure 6. Two integrals are required to calculate the area of this region.[\/caption]\n\n<p>[reveal-answer q=\"fs-id1167793473492\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793473492\"]<\/p>\n<p id=\"fs-id1167793473492\">We need to divide the interval into two pieces. The graphs of the functions intersect at [latex]x=1[\/latex] (set [latex]f(x)=g(x)[\/latex] and solve for [latex]x[\/latex]), so we evaluate two separate integrals: one over the interval [latex]\\left[0,1\\right][\/latex] and one over the interval [latex]\\left[1,2\\right].[\/latex]<\/p>\n<p id=\"fs-id1167793615262\">Over the interval [latex]\\left[0,1\\right],[\/latex] the region is bounded above by [latex]f(x)={x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis, so we have:<\/p>\n<div id=\"fs-id1167793564130\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{1}={\\displaystyle\\int }_{0}^{1}{x}^{2}dx={\\frac{{x}^{3}}{3}|}_{0}^{1}=\\frac{1}{3}.[\/latex]<\/div>\n<p id=\"fs-id1167793276900\">Over the interval [latex]\\left[1,2\\right],[\/latex] the region is bounded above by [latex]g(x)=2-x[\/latex] and below by the [latex]x\\text{-axis,}[\/latex] so we have:<\/p>\n<div id=\"fs-id1167793372769\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{2}={\\displaystyle\\int }_{1}^{2}(2-x)dx={\\left[2x-\\frac{{x}^{2}}{2}\\right]|}_{1}^{2}=\\frac{1}{2}.[\/latex]<\/div>\n<p id=\"fs-id1167793619930\">Adding these areas together, we obtain:<\/p>\n<div id=\"fs-id1167793619934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={A}_{1}+{A}_{2}=\\frac{1}{3}+\\frac{1}{2}=\\frac{5}{6}.[\/latex]<\/div>\n<p id=\"fs-id1167794209440\">The area of the region is [latex]\\frac{5}{6}[\/latex] units<sup>2<\/sup>.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find the area of the region bounded by [latex]y = x^2[\/latex] and [latex]y = 4 - x^2[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><br>\n[reveal-answer q=\"397213\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"397213\"]<\/p>\n<p>Find intersection points:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nx^2 &amp;=&amp; 4 - x^2 \\\\<br>\n2x^2 &amp;=&amp; 4 \\\\<br>\nx &amp;=&amp; \\pm \\sqrt{2}<br>\n\\end{array}[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]A = \\int_{-\\sqrt{2}}^{\\sqrt{2}} |4 - x^2 - x^2| dx = \\int_{-\\sqrt{2}}^{\\sqrt{2}} |4 - 2x^2| dx[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nA &amp;=&amp; \\int_{-\\sqrt{2}}^{\\sqrt{2}} (4 - 2x^2) dx\\\\<br>\n&amp;=&amp; [4x - \\frac{2}{3}x^3]_{-\\sqrt{2}}^{\\sqrt{2}} \\\\<br>\n&amp;=&amp; (4\\sqrt{2} - \\frac{2}{3}(\\sqrt{2})^3) - (-4\\sqrt{2} - \\frac{2}{3}(-\\sqrt{2})^3) \\\\<br>\n&amp;=&amp; 8\\sqrt{2} - \\frac{4}{3}(\\sqrt{2})^3 \\\\<br>\n&amp;=&amp; \\frac{16\\sqrt{2}}{3}<br>\n\\end{array}[\/latex]<\/p>\n<p>Therefore, the area of the region is [latex]\\frac{16\\sqrt{2}}{3}[\/latex] square units.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\n<\/section>\n<h2>Regions Defined with Respect to [latex]y[\/latex]<\/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\">Alternative approach to finding areas between curves by integrating with respect to [latex]y[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Useful when curves are more easily expressed as functions of [latex]y[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Can simplify calculations by requiring only one integral instead of two<\/li>\n\t<li class=\"whitespace-normal break-words\">Key formula: [latex]A = \\int_c^d [u(y) - v(y)] dy[\/latex], where [latex]u(y)[\/latex] is the right boundary and [latex]v(y)[\/latex] is the left boundary<\/li>\n\t<li class=\"whitespace-normal break-words\">Process:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Express curves as functions of [latex]y[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Determine the limits of integration ([latex]y[\/latex]-coordinates of intersection points)<\/li>\n\t<li class=\"whitespace-normal break-words\">Set up and evaluate the integral<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find the area of the region bounded by [latex]y = x^2[\/latex] and [latex]y = 2x[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><br>\n[reveal-answer q=\"476820\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"476820\"]<\/p>\n<p>Express curves as functions of [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]x = \\sqrt{y}[\/latex] and [latex]x = \\frac{y}{2}[\/latex]<\/p>\n<p>Find intersection points:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{y} = \\frac{y}{2}[\/latex]<br>\n[latex]y = 0[\/latex] or [latex]y = 4[\/latex]<\/p>\n<p>Set up and evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nA &amp;=&amp; \\int_0^4 [\\frac{y}{2} - \\sqrt{y}] dy \\\\<br>\n&amp;=&amp; [\\frac{y^2}{4} - \\frac{2y^{3\/2}}{3}]_0^4 \\\\<br>\n&amp;=&amp; (4 - \\frac{16}{3}) - (0 - 0) \\\\<br>\n&amp;=&amp; \\frac{4}{3}<br>\n\\end{array}[\/latex]<\/p>\n<p>The area is [latex]\\frac{4}{3}[\/latex] square units.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the area of the region in the first quadrant bounded by [latex]x^2 + y^2 = 4[\/latex] and [latex]y = x[\/latex].<\/p>\n<p><br>\n[reveal-answer q=\"649111\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"649111\"]<\/p>\n<p>Express curves as functions of [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]x = \\sqrt{4-y^2}[\/latex] and [latex]x = y[\/latex]<\/p>\n<p>Find intersection points:<\/p>\n<p style=\"text-align: center;\">[latex]y = \\sqrt{4-y^2}[\/latex]<br>\n[latex]y^2 = 2[\/latex], so [latex]y = \\sqrt{2}[\/latex]<\/p>\n<p>Set up and evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\nA &amp;=&amp; \\int_0^{\\sqrt{2}} [\\sqrt{4-y^2} - y] dy \\\\<br>\n&amp;=&amp; [2\\arcsin(\\frac{y}{2}) + \\frac{y}{2}\\sqrt{4-y^2} - \\frac{y^2}{2}]_0^{\\sqrt{2}} \\\\<br>\n&amp;=&amp; [2\\arcsin(\\frac{\\sqrt{2}}{2}) + 1 - 1] - [0 + 0 - 0] \\\\<br>\n&amp;=&amp; \\frac{\\pi}{2}<br>\n\\end{array}[\/latex]<\/p>\n<p>The area is [latex]\\frac{\\pi}{2} -[\/latex] square units.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate the area between two curves by integrating with respect to [latex]x[\/latex]<\/li>\n<li>Calculate the area of a compound region<\/li>\n<li>Calculate the area between two curves by integrating with respect to [latex]y[\/latex]<\/li>\n<li>Determine the most effective variable, [latex]x[\/latex] or [latex]y[\/latex], for integration based on the curves\u2019 orientation<\/li>\n<\/ul>\n<\/section>\n<h2>Area of a Region between Two Curves<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Extends the concept of definite integrals to calculate areas of more complex regions<\/li>\n<li class=\"whitespace-normal break-words\">For continuous functions [latex]f(x)[\/latex] and [latex]g(x)[\/latex] where f(x) \u2265 g(x) on [latex][a,b][\/latex]:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]A = \\int_a^b [f(x) - g(x)] dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Area represents the difference between the upper and lower function, integrated over the interval<\/li>\n<li class=\"whitespace-normal break-words\">Process:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Identify the upper and lower functions<\/li>\n<li class=\"whitespace-normal break-words\">Determine the interval of integration<\/li>\n<li class=\"whitespace-normal break-words\">Set up and evaluate the definite integral<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Often necessary to solve [latex]f(x) = g(x)[\/latex] to determine integration limits<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>If [latex]R[\/latex] is the region bounded by the graphs of the functions [latex]f(x)=\\frac{x}{2}+5[\/latex] and [latex]g(x)=x+\\frac{1}{2}[\/latex] over the interval [latex]\\left[1,5\\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-id1167793933114\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793933114\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793933114\">[latex]12[\/latex] units<sup>2<\/sup><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>If [latex]R[\/latex] is the region bounded above by the graph of the function [latex]f(x)=x[\/latex] and below by the graph of the function [latex]g(x)={x}^{4},[\/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-id1167793276952\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793276952\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793276952\">[latex]\\frac{3}{10}[\/latex] unit<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=616&amp;end=707&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\/2.1AreaBetweenCurves616to707_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<h2>Areas of Compound Regions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Extends area calculations to regions where functions intersect<\/li>\n<li class=\"whitespace-normal break-words\">Key formula for regions bounded by intersecting curves [latex]f(x)[\/latex] and [latex]g(x)[\/latex] on [latex][a,b][\/latex]:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]A = \\int_a^b |f(x) - g(x)| dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Absolute value in the formula accounts for changes in which function is greater<\/li>\n<li class=\"whitespace-normal break-words\">Process:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Identify intersection points of the functions<\/li>\n<li class=\"whitespace-normal break-words\">Break the interval into subintervals where one function is consistently greater<\/li>\n<li class=\"whitespace-normal break-words\">Set up and evaluate separate integrals for each subinterval<\/li>\n<li class=\"whitespace-normal break-words\">Sum the results to get the total area<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\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[\\frac{\\pi}{2},2\\pi \\right],[\/latex] find the area of region [latex]R.[\/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=\"q118401\">Hint<\/button><\/p>\n<div id=\"q118401\" class=\"hidden-answer\" style=\"display: none\">The two curves intersect at [latex]x=\\frac{(5\\pi )}{4}.[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793932268\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793932268\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793932268\">[latex]2+2\\sqrt{2}[\/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=1007&amp;end=1168&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\/2.1AreaBetweenCurves1007to1168_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<section class=\"textbox example\">\n<p id=\"fs-id1167793517455\">Consider the region depicted below. Find the area of [latex]R.[\/latex]<\/p>\n<figure style=\"width: 229px\" 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\/11212623\/CNX_Calc_Figure_06_01_006.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = x^2 and g(x)= 2-x. 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 region is labeled R. The shaded area is between x=0 and x=2.\" width=\"229\" height=\"242\" \/><figcaption class=\"wp-caption-text\">Figure 6. Two integrals are required to calculate the area of this region.<\/figcaption><\/figure>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793473492\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793473492\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793473492\">We need to divide the interval into two pieces. The graphs of the functions intersect at [latex]x=1[\/latex] (set [latex]f(x)=g(x)[\/latex] and solve for [latex]x[\/latex]), so we evaluate two separate integrals: one over the interval [latex]\\left[0,1\\right][\/latex] and one over the interval [latex]\\left[1,2\\right].[\/latex]<\/p>\n<p id=\"fs-id1167793615262\">Over the interval [latex]\\left[0,1\\right],[\/latex] the region is bounded above by [latex]f(x)={x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis, so we have:<\/p>\n<div id=\"fs-id1167793564130\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{1}={\\displaystyle\\int }_{0}^{1}{x}^{2}dx={\\frac{{x}^{3}}{3}|}_{0}^{1}=\\frac{1}{3}.[\/latex]<\/div>\n<p id=\"fs-id1167793276900\">Over the interval [latex]\\left[1,2\\right],[\/latex] the region is bounded above by [latex]g(x)=2-x[\/latex] and below by the [latex]x\\text{-axis,}[\/latex] so we have:<\/p>\n<div id=\"fs-id1167793372769\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{2}={\\displaystyle\\int }_{1}^{2}(2-x)dx={\\left[2x-\\frac{{x}^{2}}{2}\\right]|}_{1}^{2}=\\frac{1}{2}.[\/latex]<\/div>\n<p id=\"fs-id1167793619930\">Adding these areas together, we obtain:<\/p>\n<div id=\"fs-id1167793619934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={A}_{1}+{A}_{2}=\\frac{1}{3}+\\frac{1}{2}=\\frac{5}{6}.[\/latex]<\/div>\n<p id=\"fs-id1167794209440\">The area of the region is [latex]\\frac{5}{6}[\/latex] units<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find the area of the region bounded by [latex]y = x^2[\/latex] and [latex]y = 4 - x^2[\/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=\"q397213\">Show Answer<\/button><\/p>\n<div id=\"q397213\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find intersection points:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> x^2 &=& 4 - x^2 \\\\<br \/> 2x^2 &=& 4 \\\\<br \/> x &=& \\pm \\sqrt{2}<br \/> \\end{array}[\/latex]<\/p>\n<p>Set up the integral:<\/p>\n<p style=\"text-align: center;\">[latex]A = \\int_{-\\sqrt{2}}^{\\sqrt{2}} |4 - x^2 - x^2| dx = \\int_{-\\sqrt{2}}^{\\sqrt{2}} |4 - 2x^2| dx[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> A &=& \\int_{-\\sqrt{2}}^{\\sqrt{2}} (4 - 2x^2) dx\\\\<br \/> &=& [4x - \\frac{2}{3}x^3]_{-\\sqrt{2}}^{\\sqrt{2}} \\\\<br \/> &=& (4\\sqrt{2} - \\frac{2}{3}(\\sqrt{2})^3) - (-4\\sqrt{2} - \\frac{2}{3}(-\\sqrt{2})^3) \\\\<br \/> &=& 8\\sqrt{2} - \\frac{4}{3}(\\sqrt{2})^3 \\\\<br \/> &=& \\frac{16\\sqrt{2}}{3}<br \/> \\end{array}[\/latex]<\/p>\n<p>Therefore, the area of the region is [latex]\\frac{16\\sqrt{2}}{3}[\/latex] square units.<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<h2>Regions Defined with Respect to [latex]y[\/latex]<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Alternative approach to finding areas between curves by integrating with respect to [latex]y[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Useful when curves are more easily expressed as functions of [latex]y[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Can simplify calculations by requiring only one integral instead of two<\/li>\n<li class=\"whitespace-normal break-words\">Key formula: [latex]A = \\int_c^d [u(y) - v(y)] dy[\/latex], where [latex]u(y)[\/latex] is the right boundary and [latex]v(y)[\/latex] is the left boundary<\/li>\n<li class=\"whitespace-normal break-words\">Process:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Express curves as functions of [latex]y[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Determine the limits of integration ([latex]y[\/latex]-coordinates of intersection points)<\/li>\n<li class=\"whitespace-normal break-words\">Set up and evaluate the integral<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find the area of the region bounded by [latex]y = x^2[\/latex] and [latex]y = 2x[\/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=\"q476820\">Show Answer<\/button><\/p>\n<div id=\"q476820\" class=\"hidden-answer\" style=\"display: none\">\n<p>Express curves as functions of [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]x = \\sqrt{y}[\/latex] and [latex]x = \\frac{y}{2}[\/latex]<\/p>\n<p>Find intersection points:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{y} = \\frac{y}{2}[\/latex]<br \/>\n[latex]y = 0[\/latex] or [latex]y = 4[\/latex]<\/p>\n<p>Set up and evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> A &=& \\int_0^4 [\\frac{y}{2} - \\sqrt{y}] dy \\\\<br \/> &=& [\\frac{y^2}{4} - \\frac{2y^{3\/2}}{3}]_0^4 \\\\<br \/> &=& (4 - \\frac{16}{3}) - (0 - 0) \\\\<br \/> &=& \\frac{4}{3}<br \/> \\end{array}[\/latex]<\/p>\n<p>The area is [latex]\\frac{4}{3}[\/latex] square units.<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the area of the region in the first quadrant bounded by [latex]x^2 + y^2 = 4[\/latex] and [latex]y = x[\/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=\"q649111\">Show Answer<\/button><\/p>\n<div id=\"q649111\" class=\"hidden-answer\" style=\"display: none\">\n<p>Express curves as functions of [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]x = \\sqrt{4-y^2}[\/latex] and [latex]x = y[\/latex]<\/p>\n<p>Find intersection points:<\/p>\n<p style=\"text-align: center;\">[latex]y = \\sqrt{4-y^2}[\/latex]<br \/>\n[latex]y^2 = 2[\/latex], so [latex]y = \\sqrt{2}[\/latex]<\/p>\n<p>Set up and evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> A &=& \\int_0^{\\sqrt{2}} [\\sqrt{4-y^2} - y] dy \\\\<br \/> &=& [2\\arcsin(\\frac{y}{2}) + \\frac{y}{2}\\sqrt{4-y^2} - \\frac{y^2}{2}]_0^{\\sqrt{2}} \\\\<br \/> &=& [2\\arcsin(\\frac{\\sqrt{2}}{2}) + 1 - 1] - [0 + 0 - 0] \\\\<br \/> &=& \\frac{\\pi}{2}<br \/> \\end{array}[\/latex]<\/p>\n<p>The area is [latex]\\frac{\\pi}{2} -[\/latex] square units.<\/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":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\/430"}],"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\/430\/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\/430\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=430"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=430"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=430"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=430"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}