{"id":1122,"date":"2021-06-30T17:01:56","date_gmt":"2021-06-30T17:01:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/area-and-the-definite-integral\/"},"modified":"2022-03-19T03:12:20","modified_gmt":"2022-03-19T03:12:20","slug":"area-and-the-definite-integral","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/area-and-the-definite-integral\/","title":{"raw":"Area and the Definite Integral","rendered":"Area and the Definite Integral"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the relationship between the definite integral and net area<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572430392\">When we defined the definite integral, we lifted the requirement that [latex]f(x)[\/latex] be nonnegative. But how do we interpret \u201cthe area under the curve\u201d when [latex]f(x)[\/latex] is negative?<\/p>\r\n\r\n<div id=\"fs-id1170571600645\" class=\"bc-section section\">\r\n<h2>Net Signed Area<\/h2>\r\n<p id=\"fs-id1170571600651\">Let us return to the Riemann sum. Consider, for example, the function [latex]f(x)=2-2x^2[\/latex] (shown in Figure 2) on the interval [latex][0,2][\/latex]. Use [latex]n=8[\/latex] and choose [latex]x_i^*[\/latex] as the left endpoint of each interval. Construct a rectangle on each subinterval of height [latex]f(x_i^*)[\/latex] and width [latex]\\Delta x[\/latex]. When [latex]f(x_i^*)[\/latex] is positive, the product [latex]f(x_i^*) \\Delta x[\/latex] represents the area of the rectangle, as before. When [latex]f(x_i^*)[\/latex] is negative, however, the product [latex]f(x_i^*) \\Delta x[\/latex] represents the <em>negative<\/em> of the area of the rectangle. The Riemann sum then becomes<\/p>\r\n\r\n<div id=\"fs-id1170572409521\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{i=1}^{8}f(x_i^*) \\Delta x =[\/latex] (Area of rectangles above the [latex]x[\/latex]-axis) [latex]-[\/latex] (Area of rectangles below the [latex]x[\/latex]-axis)<\/div>\r\n&nbsp;\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204006\/CNX_Calc_Figure_05_02_003.jpg\" alt=\"A graph of a downward opening parabola over [-1, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each.\" width=\"325\" height=\"387\" \/> Figure 2. For a function that is partly negative, the Riemann sum is the area of the rectangles above the [latex]x[\/latex]-axis minus the area of the rectangles below the [latex]x[\/latex]-axis.[\/caption]<\/div>\r\n<p id=\"fs-id1170571671431\">Taking the limit as [latex]n\\to \\infty[\/latex], the Riemann sum approaches the area between the curve above the [latex]x[\/latex]-axis and the [latex]x[\/latex]-axis, minus the area between the curve below the [latex]x[\/latex]-axis and the [latex]x[\/latex]-axis, as shown in Figure 3. Then,<\/p>\r\n\r\n<div id=\"fs-id1170571671468\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 f(x) dx &amp; =\\underset{n\\to \\infty}{\\lim} \\displaystyle\\sum_{i=1}^{n} f(c_i) \\Delta x \\\\ &amp; =A_1-A_2 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572294442\">The quantity [latex]A_1-A_2[\/latex] is called the <strong>net signed area<\/strong>.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204010\/CNX_Calc_Figure_05_02_002.jpg\" alt=\"A graph of a downward opening parabola over [-2, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). The area in quadrant one under the curve is shaded blue and labeled A1. The area in quadrant four above the curve and to the left of x=2 is shaded blue and labeled A2.\" width=\"325\" height=\"350\" \/> Figure 3. In the limit, the definite integral equals area [latex]A_1[\/latex] minus area [latex]A_2[\/latex], or the net signed area.[\/caption]\r\n<p id=\"fs-id1170572449095\">Notice that net signed area can be positive, negative, or zero. If the area above the [latex]x[\/latex]-axis is larger, the net signed area is positive. If the area below the [latex]x[\/latex]-axis is larger, the net signed area is negative. If the areas above and below the [latex]x[\/latex]-axis are equal, the net signed area is zero.<\/p>\r\n\r\n<div id=\"fs-id1170572449111\" class=\"textbox exercises\">\r\n<h3>Example: Finding the Net Signed Area<\/h3>\r\nFind the net signed area between the curve of the function [latex]f(x)=2x[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][-3,3][\/latex].\r\n[reveal-answer q=\"fs-id1170571712242\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571712242\"]\r\n\r\nThe function produces a straight line that forms two triangles: one from [latex]x=-3[\/latex] to [latex]x=0[\/latex] and the other from [latex]x=0[\/latex] to [latex]x=3[\/latex] (Figure 4). Using the geometric formula for the area of a triangle, [latex]A=\\frac{1}{2}bh[\/latex], the area of triangle [latex]A_1[\/latex], above the axis, is\r\n<p style=\"text-align: center;\">[latex]A_1=\\frac{1}{2}3(6)=9[\/latex],<\/p>\r\n&nbsp;\r\n<p style=\"text-align: left;\">where 3 is the base and [latex]2(3)=6[\/latex] is the height. The area of triangle [latex]A_2[\/latex], below the axis, is<\/p>\r\n\r\n<div id=\"fs-id1170571609249\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_2=\\frac{1}{2}(3)(6)=9[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572380038\">where 3 is the base and 6 is the height. Thus, the net area is<\/p>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]\\displaystyle\\int_{-3}^3 2x dx=A_1-A_2=9-9=0[\/latex].<\/span><\/p>\r\n&nbsp;\r\n<div id=\"fs-id1170572380041\" class=\"equation unnumbered\" style=\"text-align: center;\">\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204013\/CNX_Calc_Figure_05_02_005.jpg\" alt=\"A graph of an increasing line over [-6, 6] going through the origin and (-3, -6) and (3,6). The area under the line in quadrant one over [0,3] is shaded blue and labeled A1, and the area above the line in quadrant three over [-3,0] is shaded blue and labeled A2.\" width=\"325\" height=\"275\" \/> Figure 4. The area above the curve and below the [latex]x[\/latex]-axis equals the area below the curve and above the [latex]x[\/latex]-axis.[\/caption]<\/div>\r\n<h4 id=\"fs-id1170571699045\">Analysis<\/h4>\r\nIf [latex]A_1[\/latex]\u00a0is the area above the [latex]x[\/latex]-axis and [latex]A_2[\/latex]\u00a0is the area below the [latex]x[\/latex]-axis, then the net area is [latex]A_1-A_2[\/latex]. Since the areas of the two triangles are equal, the net area is zero.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding the Net Signed Area.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tto0E7yOSLo?controls=0&amp;start=520&amp;end=689&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\/5.2TheDefiniteIntegral520to689_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-id1170572338454\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572338461\">Find the net signed area of [latex]f(x)=x-2[\/latex] over the interval [latex][0,6][\/latex], illustrated in the following image.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204016\/CNX_Calc_Figure_05_02_006.jpg\" alt=\"A graph of an increasing line going through (-2,-4), (0,-2), (2,0), (4,2) and (6,4). The area above the curve in quadrant four is shaded blue and labeled A2, and the area under the curve and to the left of x=6 in quadrant one is shaded and labeled A1.\" width=\"325\" height=\"238\" \/> Figure 5.[\/caption]\r\n\r\n[reveal-answer q=\"709003\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"709003\"]Use the solving method described in the last example.[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572420084\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572420084\"]\r\n<p id=\"fs-id1170572420084\">6<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Total Area<\/h2>\r\n<p id=\"fs-id1170572420097\">One application of the definite integral is finding <span class=\"no-emphasis\">displacement<\/span> when given a velocity function. If [latex]v(t)[\/latex] represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. This is a very important application of the definite integral, and we examine it in more detail later in the chapter. For now, we\u2019re just going to look at some basics to get a feel for how this works by studying constant velocities.<\/p>\r\n<p id=\"fs-id1170572624823\">When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car travels away from its starting position in a straight line at a speed of 70 mph for 2 hours, then it is 140 mi away from its original position (Figure 6). Using integral notation, we have<\/p>\r\n\r\n<div id=\"fs-id1170572624833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_0^2 70 dt=140[\/latex]<\/div>\r\n&nbsp;\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"566\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204020\/CNX_Calc_Figure_05_02_015.jpg\" alt=\"A graph in quadrant 1 with the x-axis labeled as t (hours) and y-axis labeled as v (mi\/hr). The area under the line v(t) = 75 is shaded blue over [0,2].\" width=\"566\" height=\"352\" \/> Figure 6. The area under the curve [latex]v(t)=75[\/latex] tells us how far the car is from its starting point at a given time.[\/caption]<\/div>\r\n<p id=\"fs-id1170571637505\">In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at a speed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at a speed of 40 mph for 3 hours, it will be back at it starting position (Figure 7). Again, using integral notation, we have<\/p>\r\n\r\n<div id=\"fs-id1170571637515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 60 dt + \\displaystyle\\int_2^5 -40 dt &amp; =120-120 \\\\ &amp; =0 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571609297\">In this case the displacement is zero.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"521\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204023\/CNX_Calc_Figure_05_02_016.jpg\" alt=\"A graph in quadrants one and four with the x-axis labeled as t (hours) and the y-axis labeled as v (mi\/hr). The first part of the graph is the line v(t) = 60 over [0,2], and the area under the line in quadrant one is shaded. The second part of the graph is the line v(t) = -40 over [2,5], and the area above the line in quadrant four is shaded.\" width=\"521\" height=\"497\" \/> Figure 7. The area above the axis and the area below the axis are equal, so the net signed area is zero.[\/caption]\r\n<p id=\"fs-id1170571609320\">Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area between the curve and the [latex]x[\/latex]-axis, regardless of whether that area is above or below the axis. This is called the <strong>total area<\/strong>.<\/p>\r\n<p id=\"fs-id1170571609332\">Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis (rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is<\/p>\r\n\r\n<div id=\"fs-id1170572379484\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 |60| dt + \\displaystyle\\int_2^5 |-40| dt &amp; = \\displaystyle\\int_0^2 60 dt + \\displaystyle\\int_2^5 40 dt \\\\ &amp; =120+120 \\\\ &amp; =240 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572368682\">Bringing these ideas together formally, we state the following definitions.<\/p>\r\n\r\n<div id=\"fs-id1170572368685\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572368688\">Let [latex]f(x)[\/latex] be an<strong> integrable function<\/strong> defined on an interval [latex][a,b][\/latex]. Let [latex]A_1[\/latex]\u00a0represent the area between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis that lies <em>above<\/em> the axis and let [latex]A_2[\/latex]\u00a0represent the area between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis that lies <em>below<\/em> the axis. Then, the <strong>net signed area<\/strong> between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis is given by<\/p>\r\n\r\n<div id=\"fs-id1170572551798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b f(x) dx = A_1-A_2[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572376450\">The <strong>total area<\/strong> between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis is given by<\/p>\r\n\r\n<div id=\"fs-id1170572376474\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b |f(x)| dx = A_1+A_2[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572163843\" class=\"textbox exercises\">\r\n<h3>Example: Finding the Total Area<\/h3>\r\nFind the total area between [latex]f(x)=x-2[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][0,6][\/latex].\r\n<div id=\"fs-id1170572163845\" class=\"exercise\">[reveal-answer q=\"fs-id1170572444230\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572444230\"]\r\n<p id=\"fs-id1170572444230\">Calculate the [latex]x[\/latex]-intercept as [latex](2,0)[\/latex] (set [latex]y=0[\/latex], solve for [latex]x[\/latex]). To find the total area, take the area below the [latex]x[\/latex]-axis over the subinterval [latex][0,2][\/latex] and add it to the area above the [latex]x[\/latex]-axis on the subinterval [latex][2,6][\/latex] (Figure 8).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204026\/CNX_Calc_Figure_05_02_008.jpg\" alt=\"A graph of a increasing line f(x) = x-2 going through the points (-2,-4), (0,2), (2,0), (4,2), and (6,4). The area under the line in quadrant one and to the left of the line x=6 is shaded and labeled A1. The area above the line in quadrant four is shaded and labeled A2.\" width=\"325\" height=\"238\" \/> Figure 8. The total area between the line and the [latex]x[\/latex]-axis over [latex][0,6][\/latex] is [latex]A_2[\/latex] plus [latex]A_1[\/latex].[\/caption]\r\n<p id=\"fs-id1170571599762\">We have<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_0^6 |(x-2)| dx = A_2+A_1[\/latex]<\/div>\r\n<p id=\"fs-id1170572229842\">Then, using the formula for the area of a triangle, we obtain<\/p>\r\n\r\n<div id=\"fs-id1170572229845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_2=\\frac{1}{2}bh=\\frac{1}{2} \\cdot 2 \\cdot 2=2[\/latex]<\/div>\r\n<div id=\"fs-id1170572217489\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_1=\\frac{1}{2}bh=\\frac{1}{2} \\cdot 4 \\cdot 4=8[\/latex]<\/div>\r\n<p id=\"fs-id1170572621626\">The total area, then, is<\/p>\r\n\r\n<div id=\"fs-id1170572621629\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_1+A_2=8+2=10[\/latex].<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding the Total Area.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tto0E7yOSLo?controls=0&amp;start=694&amp;end=817&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.2TheDefiniteIntegral694to817_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-id1170572621667\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572621674\">Find the total area between the function [latex]f(x)=2x[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][-3,3][\/latex].<\/p>\r\n[reveal-answer q=\"743821\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"743821\"]Review the solving strategy in the previous example.[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571711340\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571711340\"]18\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the relationship between the definite integral and net area<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572430392\">When we defined the definite integral, we lifted the requirement that [latex]f(x)[\/latex] be nonnegative. But how do we interpret \u201cthe area under the curve\u201d when [latex]f(x)[\/latex] is negative?<\/p>\n<div id=\"fs-id1170571600645\" class=\"bc-section section\">\n<h2>Net Signed Area<\/h2>\n<p id=\"fs-id1170571600651\">Let us return to the Riemann sum. Consider, for example, the function [latex]f(x)=2-2x^2[\/latex] (shown in Figure 2) on the interval [latex][0,2][\/latex]. Use [latex]n=8[\/latex] and choose [latex]x_i^*[\/latex] as the left endpoint of each interval. Construct a rectangle on each subinterval of height [latex]f(x_i^*)[\/latex] and width [latex]\\Delta x[\/latex]. When [latex]f(x_i^*)[\/latex] is positive, the product [latex]f(x_i^*) \\Delta x[\/latex] represents the area of the rectangle, as before. When [latex]f(x_i^*)[\/latex] is negative, however, the product [latex]f(x_i^*) \\Delta x[\/latex] represents the <em>negative<\/em> of the area of the rectangle. The Riemann sum then becomes<\/p>\n<div id=\"fs-id1170572409521\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{i=1}^{8}f(x_i^*) \\Delta x =[\/latex] (Area of rectangles above the [latex]x[\/latex]-axis) [latex]-[\/latex] (Area of rectangles below the [latex]x[\/latex]-axis)<\/div>\n<p>&nbsp;<\/p>\n<div>\n<div style=\"width: 335px\" 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\/11204006\/CNX_Calc_Figure_05_02_003.jpg\" alt=\"A graph of a downward opening parabola over [-1, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each.\" width=\"325\" height=\"387\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. For a function that is partly negative, the Riemann sum is the area of the rectangles above the [latex]x[\/latex]-axis minus the area of the rectangles below the [latex]x[\/latex]-axis.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571671431\">Taking the limit as [latex]n\\to \\infty[\/latex], the Riemann sum approaches the area between the curve above the [latex]x[\/latex]-axis and the [latex]x[\/latex]-axis, minus the area between the curve below the [latex]x[\/latex]-axis and the [latex]x[\/latex]-axis, as shown in Figure 3. Then,<\/p>\n<div id=\"fs-id1170571671468\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 f(x) dx & =\\underset{n\\to \\infty}{\\lim} \\displaystyle\\sum_{i=1}^{n} f(c_i) \\Delta x \\\\ & =A_1-A_2 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572294442\">The quantity [latex]A_1-A_2[\/latex] is called the <strong>net signed area<\/strong>.<\/p>\n<div style=\"width: 335px\" 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\/11204010\/CNX_Calc_Figure_05_02_002.jpg\" alt=\"A graph of a downward opening parabola over [-2, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). The area in quadrant one under the curve is shaded blue and labeled A1. The area in quadrant four above the curve and to the left of x=2 is shaded blue and labeled A2.\" width=\"325\" height=\"350\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. In the limit, the definite integral equals area [latex]A_1[\/latex] minus area [latex]A_2[\/latex], or the net signed area.<\/p>\n<\/div>\n<p id=\"fs-id1170572449095\">Notice that net signed area can be positive, negative, or zero. If the area above the [latex]x[\/latex]-axis is larger, the net signed area is positive. If the area below the [latex]x[\/latex]-axis is larger, the net signed area is negative. If the areas above and below the [latex]x[\/latex]-axis are equal, the net signed area is zero.<\/p>\n<div id=\"fs-id1170572449111\" class=\"textbox exercises\">\n<h3>Example: Finding the Net Signed Area<\/h3>\n<p>Find the net signed area between the curve of the function [latex]f(x)=2x[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][-3,3][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571712242\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571712242\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function produces a straight line that forms two triangles: one from [latex]x=-3[\/latex] to [latex]x=0[\/latex] and the other from [latex]x=0[\/latex] to [latex]x=3[\/latex] (Figure 4). Using the geometric formula for the area of a triangle, [latex]A=\\frac{1}{2}bh[\/latex], the area of triangle [latex]A_1[\/latex], above the axis, is<\/p>\n<p style=\"text-align: center;\">[latex]A_1=\\frac{1}{2}3(6)=9[\/latex],<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\">where 3 is the base and [latex]2(3)=6[\/latex] is the height. The area of triangle [latex]A_2[\/latex], below the axis, is<\/p>\n<div id=\"fs-id1170571609249\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_2=\\frac{1}{2}(3)(6)=9[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572380038\">where 3 is the base and 6 is the height. Thus, the net area is<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]\\displaystyle\\int_{-3}^3 2x dx=A_1-A_2=9-9=0[\/latex].<\/span><\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1170572380041\" class=\"equation unnumbered\" style=\"text-align: center;\">\n<div>\n<div style=\"width: 335px\" 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\/11204013\/CNX_Calc_Figure_05_02_005.jpg\" alt=\"A graph of an increasing line over &#091;-6, 6&#093; going through the origin and (-3, -6) and (3,6). The area under the line in quadrant one over &#091;0,3&#093; is shaded blue and labeled A1, and the area above the line in quadrant three over &#091;-3,0&#093; is shaded blue and labeled A2.\" width=\"325\" height=\"275\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. The area above the curve and below the [latex]x[\/latex]-axis equals the area below the curve and above the [latex]x[\/latex]-axis.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-id1170571699045\">Analysis<\/h4>\n<p>If [latex]A_1[\/latex]\u00a0is the area above the [latex]x[\/latex]-axis and [latex]A_2[\/latex]\u00a0is the area below the [latex]x[\/latex]-axis, then the net area is [latex]A_1-A_2[\/latex]. Since the areas of the two triangles are equal, the net area is zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Net Signed Area.<\/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=520&amp;end=689&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\/5.2TheDefiniteIntegral520to689_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-id1170572338454\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572338461\">Find the net signed area of [latex]f(x)=x-2[\/latex] over the interval [latex][0,6][\/latex], illustrated in the following image.<\/p>\n<div style=\"width: 335px\" 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\/11204016\/CNX_Calc_Figure_05_02_006.jpg\" alt=\"A graph of an increasing line going through (-2,-4), (0,-2), (2,0), (4,2) and (6,4). The area above the curve in quadrant four is shaded blue and labeled A2, and the area under the curve and to the left of x=6 in quadrant one is shaded and labeled A1.\" width=\"325\" height=\"238\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q709003\">Hint<\/span><\/p>\n<div id=\"q709003\" class=\"hidden-answer\" style=\"display: none\">Use the solving method described in the last example.<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572420084\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572420084\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572420084\">6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Total Area<\/h2>\n<p id=\"fs-id1170572420097\">One application of the definite integral is finding <span class=\"no-emphasis\">displacement<\/span> when given a velocity function. If [latex]v(t)[\/latex] represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. This is a very important application of the definite integral, and we examine it in more detail later in the chapter. For now, we\u2019re just going to look at some basics to get a feel for how this works by studying constant velocities.<\/p>\n<p id=\"fs-id1170572624823\">When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car travels away from its starting position in a straight line at a speed of 70 mph for 2 hours, then it is 140 mi away from its original position (Figure 6). Using integral notation, we have<\/p>\n<div id=\"fs-id1170572624833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_0^2 70 dt=140[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div>\n<div style=\"width: 576px\" 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\/11204020\/CNX_Calc_Figure_05_02_015.jpg\" alt=\"A graph in quadrant 1 with the x-axis labeled as t (hours) and y-axis labeled as v (mi\/hr). The area under the line v(t) = 75 is shaded blue over [0,2].\" width=\"566\" height=\"352\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. The area under the curve [latex]v(t)=75[\/latex] tells us how far the car is from its starting point at a given time.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571637505\">In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at a speed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at a speed of 40 mph for 3 hours, it will be back at it starting position (Figure 7). Again, using integral notation, we have<\/p>\n<div id=\"fs-id1170571637515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 60 dt + \\displaystyle\\int_2^5 -40 dt & =120-120 \\\\ & =0 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571609297\">In this case the displacement is zero.<\/p>\n<div style=\"width: 531px\" 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\/11204023\/CNX_Calc_Figure_05_02_016.jpg\" alt=\"A graph in quadrants one and four with the x-axis labeled as t (hours) and the y-axis labeled as v (mi\/hr). The first part of the graph is the line v(t) = 60 over [0,2], and the area under the line in quadrant one is shaded. The second part of the graph is the line v(t) = -40 over [2,5], and the area above the line in quadrant four is shaded.\" width=\"521\" height=\"497\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. The area above the axis and the area below the axis are equal, so the net signed area is zero.<\/p>\n<\/div>\n<p id=\"fs-id1170571609320\">Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area between the curve and the [latex]x[\/latex]-axis, regardless of whether that area is above or below the axis. This is called the <strong>total area<\/strong>.<\/p>\n<p id=\"fs-id1170571609332\">Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis (rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is<\/p>\n<div id=\"fs-id1170572379484\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 |60| dt + \\displaystyle\\int_2^5 |-40| dt & = \\displaystyle\\int_0^2 60 dt + \\displaystyle\\int_2^5 40 dt \\\\ & =120+120 \\\\ & =240 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572368682\">Bringing these ideas together formally, we state the following definitions.<\/p>\n<div id=\"fs-id1170572368685\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170572368688\">Let [latex]f(x)[\/latex] be an<strong> integrable function<\/strong> defined on an interval [latex][a,b][\/latex]. Let [latex]A_1[\/latex]\u00a0represent the area between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis that lies <em>above<\/em> the axis and let [latex]A_2[\/latex]\u00a0represent the area between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis that lies <em>below<\/em> the axis. Then, the <strong>net signed area<\/strong> between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis is given by<\/p>\n<div id=\"fs-id1170572551798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b f(x) dx = A_1-A_2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572376450\">The <strong>total area<\/strong> between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis is given by<\/p>\n<div id=\"fs-id1170572376474\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b |f(x)| dx = A_1+A_2[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1170572163843\" class=\"textbox exercises\">\n<h3>Example: Finding the Total Area<\/h3>\n<p>Find the total area between [latex]f(x)=x-2[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][0,6][\/latex].<\/p>\n<div id=\"fs-id1170572163845\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572444230\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572444230\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572444230\">Calculate the [latex]x[\/latex]-intercept as [latex](2,0)[\/latex] (set [latex]y=0[\/latex], solve for [latex]x[\/latex]). To find the total area, take the area below the [latex]x[\/latex]-axis over the subinterval [latex][0,2][\/latex] and add it to the area above the [latex]x[\/latex]-axis on the subinterval [latex][2,6][\/latex] (Figure 8).<\/p>\n<div style=\"width: 335px\" 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\/11204026\/CNX_Calc_Figure_05_02_008.jpg\" alt=\"A graph of a increasing line f(x) = x-2 going through the points (-2,-4), (0,2), (2,0), (4,2), and (6,4). The area under the line in quadrant one and to the left of the line x=6 is shaded and labeled A1. The area above the line in quadrant four is shaded and labeled A2.\" width=\"325\" height=\"238\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. The total area between the line and the [latex]x[\/latex]-axis over [latex][0,6][\/latex] is [latex]A_2[\/latex] plus [latex]A_1[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170571599762\">We have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_0^6 |(x-2)| dx = A_2+A_1[\/latex]<\/div>\n<p id=\"fs-id1170572229842\">Then, using the formula for the area of a triangle, we obtain<\/p>\n<div id=\"fs-id1170572229845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_2=\\frac{1}{2}bh=\\frac{1}{2} \\cdot 2 \\cdot 2=2[\/latex]<\/div>\n<div id=\"fs-id1170572217489\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_1=\\frac{1}{2}bh=\\frac{1}{2} \\cdot 4 \\cdot 4=8[\/latex]<\/div>\n<p id=\"fs-id1170572621626\">The total area, then, is<\/p>\n<div id=\"fs-id1170572621629\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A_1+A_2=8+2=10[\/latex].<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Total Area.<\/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=694&amp;end=817&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.2TheDefiniteIntegral694to817_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-id1170572621667\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572621674\">Find the total area between the function [latex]f(x)=2x[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][-3,3][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q743821\">Hint<\/span><\/p>\n<div id=\"q743821\" class=\"hidden-answer\" style=\"display: none\">Review the solving strategy in the previous example.<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571711340\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571711340\" class=\"hidden-answer\" style=\"display: none\">18<\/p>\n<\/div>\n<\/div>\n<\/div>\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-1122\">\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":9,"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-1122","chapter","type-chapter","status-publish","hentry"],"part":1113,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1122","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1122\/revisions"}],"predecessor-version":[{"id":2638,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1122\/revisions\/2638"}],"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\/1122\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1122"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1122"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1122"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1122"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}