{"id":547,"date":"2021-02-04T16:11:07","date_gmt":"2021-02-04T16:11:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=547"},"modified":"2022-03-16T22:20:27","modified_gmt":"2022-03-16T22:20:27","slug":"net-change-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/net-change-theorem\/","title":{"raw":"Net Change Theorem","rendered":"Net Change Theorem"},"content":{"raw":"<div id=\"fs-id1170572280538\" class=\"bc-section section\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply the basic integration formulas<\/li>\r\n \t<li>Explain the significance of the net change theorem<\/li>\r\n \t<li><span class=\"os-abstract-content\">Use the net change theorem to solve applied problems<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Basic Integration Formulas<\/h2>\r\n<p id=\"fs-id1170572398094\">Recall the integration formulas given in the Table of Antiderivatives and the rule on properties of definite integrals. Let\u2019s look at a few examples of how to apply these rules.<\/p>\r\n\r\n<div id=\"fs-id1170571609338\" class=\"textbox exercises\">\r\n<h3>Example: Integrating a Function Using the Power Rule<\/h3>\r\nUse the power rule to integrate the function [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt.[\/latex]\r\n<div id=\"fs-id1170572506475\" class=\"exercise\">[reveal-answer q=\"fs-id1170572431500\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572431500\"]\r\n<p id=\"fs-id1170572431500\">The first step is to rewrite the function and simplify it so we can apply the power rule:<\/p>\r\n\r\n<div id=\"fs-id1170572506256\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt\\hfill &amp; ={\\displaystyle\\int }_{1}^{4}{t}^{1\\text{\/}2}(1+t)dt\\hfill \\\\ \\\\ &amp; ={\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572150489\">Now apply the power rule:<\/p>\r\n\r\n<div id=\"fs-id1170572539674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt\\hfill &amp; ={(\\frac{2}{3}{t}^{3\\text{\/}2}+\\frac{2}{5}{t}^{5\\text{\/}2})|}_{1}^{4}\\hfill \\\\ &amp; =\\left[\\frac{2}{3}{(4)}^{3\\text{\/}2}+\\frac{2}{5}{(4)}^{5\\text{\/}2}\\right]-\\left[\\frac{2}{3}{(1)}^{3\\text{\/}2}+\\frac{2}{5}{(1)}^{5\\text{\/}2}\\right]\\hfill \\\\ &amp; =\\frac{256}{15}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Integrating a Function Using the Power Rule.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/v7nDnOyx8Mw?controls=0&amp;start=9&amp;end=96&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\/5.4IntegrationFormulasAndTheNetChangeTheorem9to96_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.4 Integration Formulas and the Net Change Theorem\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572558558\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572150887\">Find the definite integral of [latex]f(x)={x}^{2}-3x[\/latex] over the interval [latex]\\left[1,3\\right].[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572100050\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572100050\"]\r\n<p id=\"fs-id1170572100050\">[latex]-\\frac{10}{3}[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170571814551\">Follow the process from <a class=\"autogenerated-content\" href=\"#fs-id1170571609338\">(Figure)<\/a> to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20040[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572627233\" class=\"bc-section section\">\r\n<h2>The Net Change Theorem<\/h2>\r\n<p id=\"fs-id1170571583276\">The <strong>net change theorem<\/strong> considers the integral of a <span class=\"no-emphasis\"><em>rate of change<\/em><\/span>. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.<\/p>\r\n\r\n<div id=\"fs-id1170572295126\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Net Change Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572480924\">The new value of a changing quantity equals the initial value plus the integral of the rate of change:<\/p>\r\n\r\n<div id=\"fs-id1170572449540\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ F(b)=F(a)+{\\displaystyle\\int }_{a}^{b}F\\text{'}(x)dx\\hfill \\\\ \\hfill \\text{or}\\hfill \\\\ {\\displaystyle\\int }_{a}^{b}F\\text{'}(x)dx=F(b)-F(a).\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170572110596\">Subtracting [latex]F(a)[\/latex] from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.<\/p>\r\n<p id=\"fs-id1170572481528\">The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let\u2019s apply the net change theorem to a <span class=\"no-emphasis\">velocity<\/span> function in which the result is <span class=\"no-emphasis\">displacement<\/span>.<\/p>\r\nWe looked at a simple example of this in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/the-definite-integral\/\">The Definite Integral<\/a>. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_05_04_001\">(Figure)<\/a>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"286\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204144\/CNX_Calc_Figure_05_04_002.jpg\" alt=\"A graph with the x axis marked as t and the y axis marked normally. The lines y=40 and y=-30 are drawn over [2,4] and [4,5], respectively.The areas between the lines and the x axis are shaded.\" width=\"286\" height=\"309\" \/> Figure 1. The graph shows speed versus time for the given motion of a car.[\/caption]\r\n<p id=\"fs-id1170572296931\">Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by<\/p>\r\n\r\n<div id=\"fs-id1170572371074\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{2}^{5}v(t)dt\\hfill &amp; ={\\int }_{2}^{4}40dt+{\\displaystyle\\int }_{4}^{5}-30dt\\hfill \\\\ &amp; =80-30\\hfill \\\\ &amp; =50.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572480929\">Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{2}^{5}|v(t)|dt\\hfill &amp; ={\\int }_{2}^{4}40dt+{\\displaystyle\\int }_{4}^{5}30dt\\hfill \\\\ &amp; =80+30\\hfill \\\\ &amp; =110.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571657334\">Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.<\/p>\r\n<p id=\"fs-id1170572216807\">To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.<\/p>\r\n\r\n<div id=\"fs-id1170572229863\" class=\"textbox exercises\">\r\n<h3>Example: Finding Net Displacement<\/h3>\r\nGiven a velocity function [latex]v(t)=3t-5[\/latex] (in meters per second) for a particle in motion from time [latex]t=0[\/latex] to time [latex]t=3,[\/latex] find the net displacement of the particle.\r\n<div id=\"fs-id1170571655282\" class=\"exercise\">[reveal-answer q=\"fs-id1170572455638\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572455638\"]\r\n<p id=\"fs-id1170572455638\">Applying the net change theorem, we have<\/p>\r\n\r\n<div id=\"fs-id1170572451466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\int }_{0}^{3}(3t-5)dt\\hfill &amp; =\\frac{3{t}^{2}}{2}-5t{|}_{0}^{3}\\hfill \\\\ \\\\ &amp; =\\left[\\frac{3{(3)}^{2}}{2}-5(3)\\right]-0\\hfill \\\\ &amp; =\\frac{27}{2}-15\\hfill \\\\ &amp; =\\frac{27}{2}-\\frac{30}{2}\\hfill \\\\ &amp; =-\\frac{3}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571655148\">The net displacement is [latex]-\\frac{3}{2}[\/latex] m (<a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_05_04_002\">(Figure)<\/a>).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204148\/CNX_Calc_Figure_05_04_003.jpg\" alt=\"A graph of the line v(t) = 3t \u2013 5, which goes through points (0, -5) and (5\/3, 0). The area over the line and under the x axis in the interval [0, 5\/3] is shaded. The area under the line and above the x axis in the interval [5\/3, 3] is shaded.\" width=\"304\" height=\"422\" \/> Figure 2. The graph shows velocity versus time for a particle moving with a linear velocity function.[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572247866\" class=\"textbox exercises\">\r\n<h3>Example: Finding the Total Distance Traveled<\/h3>\r\nUse <a class=\"autogenerated-content\" href=\"#fs-id1170572229863\">(Figure)<\/a> to find the total distance traveled by a particle according to the velocity function [latex]v(t)=3t-5[\/latex] m\/sec over a time interval [latex]\\left[0,3\\right].[\/latex]\r\n<div id=\"fs-id1170572510360\" class=\"exercise\">[reveal-answer q=\"fs-id1170572206288\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572206288\"]\r\n<p id=\"fs-id1170572206288\">The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.<\/p>\r\n<p id=\"fs-id1170572206422\">To continue with the example, use two integrals to find the total distance. First, find the [latex]t[\/latex]-intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for [latex]t[\/latex].<\/p>\r\nThus,\r\n<div id=\"fs-id1170571809306\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}3t-5\\hfill &amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill 3t&amp; =\\hfill &amp; 5\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{5}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572522381\">The two subintervals are [latex]\\left[0,\\frac{5}{3}\\right][\/latex] and [latex]\\left[\\frac{5}{3},3\\right].[\/latex] To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval [latex]\\left[0,\\frac{5}{3}\\right],[\/latex] we have [latex]|v(t)|=\\text{\u2212}v(t)[\/latex] over that interval. Over [latex]\\left[\\frac{5}{3},3\\right],[\/latex] the function is positive, so [latex]|v(t)|=v(t).[\/latex] Thus, we have<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\int }_{0}^{3}|v(t)|dt\\hfill &amp; ={\\int }_{0}^{5\\text{\/}3}\\text{\u2212}v(t)dt+{\\int }_{5\\text{\/}3}^{3}v(t)dt\\hfill \\\\ \\\\ &amp; ={\\int }_{0}^{5\\text{\/}3}5-3tdt+{\\int }_{5\\text{\/}3}^{3}3t-5dt\\hfill \\\\ &amp; ={(5t-\\frac{3{t}^{2}}{2})|}_{0}^{5\\text{\/}3}+{(\\frac{3{t}^{2}}{2}-5t)|}_{5\\text{\/}3}^{3}\\hfill \\\\ &amp; =\\left[5(\\frac{5}{3})-\\frac{3{(5\\text{\/}3)}^{2}}{2}\\right]-0+\\left[\\frac{27}{2}-15\\right]-\\left[\\frac{3{(5\\text{\/}3)}^{2}}{2}-\\frac{25}{3}\\right]\\hfill \\\\ &amp; =\\frac{25}{3}-\\frac{25}{6}+\\frac{27}{2}-15-\\frac{25}{6}+\\frac{25}{3}\\hfill \\\\ &amp; =\\frac{41}{6}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572624276\">So, the total distance traveled is [latex]\\frac{41}{6}[\/latex] m.<\/p>\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 Total Distance Traveled.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/v7nDnOyx8Mw?controls=0&amp;start=244&amp;end=456&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\/5.4IntegrationFormulasAndTheNetChangeTheorem244to456_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.4 Integration Formulas and the Net Change Theorem\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572622550\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571609321\">Find the net displacement and total distance traveled in meters given the velocity function [latex]f(t)=\\frac{1}{2}{e}^{t}-2[\/latex] over the interval [latex]\\left[0,2\\right].[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571637281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571637281\"]\r\n<p id=\"fs-id1170571637281\">Net displacement: [latex]\\frac{{e}^{2}-9}{2}\\approx -0.8055\\text{m;}[\/latex] total distance traveled: [latex]4\\text{ln}4-7.5+\\frac{{e}^{2}}{2}\\approx 1.740[\/latex] m<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170572244843\">Follow the procedures from <a class=\"autogenerated-content\" href=\"#fs-id1170572229863\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1170572247866\">(Figure)<\/a>. Note that [latex]f(t)\\le 0[\/latex] for [latex]t\\le \\text{ln}4,[\/latex] and [latex]f(t)\\ge 0[\/latex] for [latex]t\\ge \\text{ln}4.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572547644\" class=\"bc-section section\">\r\n<h2>Applying the Net Change Theorem<\/h2>\r\n<p id=\"fs-id1170572294455\">The net change theorem can be applied to the flow and consumption of fluids, as shown in the example below.<\/p>\r\n\r\n<div id=\"fs-id1170571542385\" class=\"textbox exercises\">\r\n<h3><span style=\"color: #000000;\">Example: How Many Gallons of Gasoline Are Consumed?<\/span><\/h3>\r\nIf the motor on a motorboat is started at [latex]t=0[\/latex] and the boat consumes gasoline at a rate which can be modeled for the first two hours as [latex]5-\\frac{t^{3}}{100}[\/latex] gal\/hr for the first hour, how much gasoline is used in the first hour?\r\n<div id=\"fs-id1170571542387\" class=\"exercise\">[reveal-answer q=\"fs-id1170572242305\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572242305\"]\r\n<p id=\"fs-id1170572242305\">Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval [latex]\\left[0,1\\right].[\/latex] We have<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int _{0}^{1}}\\left(5-\\dfrac{{t}^{3}}{100}\\right)dt\\hfill &amp; =\\left(5t-\\dfrac{{t}^{4}}{400}\\right){\\displaystyle |_{0}^{1}}\\hfill \\\\ \\\\ \\\\ &amp; =\\left[5(1)-\\dfrac{{(1)}^{4}}{400}\\right]-0\\hfill \\\\ &amp; =5-\\frac{1}{400}\\hfill \\\\ &amp; =4.9975.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572425086\">Thus, the motorboat uses 4.9975 gal of gas in 1 hour.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572379540\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170571609276\">Suppose that, instead of remaining steady during the second half hour of Andrew\u2019s outing, the wind starts to die down according to the function [latex]v(t)=-10t+15.[\/latex] In other words, the wind speed is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]v(t)=\\bigg\\{\\begin{array}{lll}20t+5\\hfill &amp; \\text{ for }\\hfill &amp; 0\\le t\\le \\frac{1}{2}\\hfill \\\\ -10t+15\\hfill &amp; \\text{ for }\\hfill &amp; \\frac{1}{2}\\le t\\le 1.\\hfill \\end{array}[\/latex]<\/p>\r\nUnder these conditions, how far from his starting point is Andrew after 1 hour?\r\n\r\n[reveal-answer q=\"39980244\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"39980244\"]\r\n<p id=\"fs-id1170572181954\">Don\u2019t forget that Andrew\u2019s iceboat moves twice as fast as the wind.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571711312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571711312\"]\r\n<p id=\"fs-id1170571711312\">17.5 mi<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>","rendered":"<div id=\"fs-id1170572280538\" class=\"bc-section section\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply the basic integration formulas<\/li>\n<li>Explain the significance of the net change theorem<\/li>\n<li><span class=\"os-abstract-content\">Use the net change theorem to solve applied problems<\/span><\/li>\n<\/ul>\n<\/div>\n<h2>Basic Integration Formulas<\/h2>\n<p id=\"fs-id1170572398094\">Recall the integration formulas given in the Table of Antiderivatives and the rule on properties of definite integrals. Let\u2019s look at a few examples of how to apply these rules.<\/p>\n<div id=\"fs-id1170571609338\" class=\"textbox exercises\">\n<h3>Example: Integrating a Function Using the Power Rule<\/h3>\n<p>Use the power rule to integrate the function [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt.[\/latex]<\/p>\n<div id=\"fs-id1170572506475\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572431500\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572431500\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572431500\">The first step is to rewrite the function and simplify it so we can apply the power rule:<\/p>\n<div id=\"fs-id1170572506256\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt\\hfill & ={\\displaystyle\\int }_{1}^{4}{t}^{1\\text{\/}2}(1+t)dt\\hfill \\\\ \\\\ & ={\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572150489\">Now apply the power rule:<\/p>\n<div id=\"fs-id1170572539674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt\\hfill & ={(\\frac{2}{3}{t}^{3\\text{\/}2}+\\frac{2}{5}{t}^{5\\text{\/}2})|}_{1}^{4}\\hfill \\\\ & =\\left[\\frac{2}{3}{(4)}^{3\\text{\/}2}+\\frac{2}{5}{(4)}^{5\\text{\/}2}\\right]-\\left[\\frac{2}{3}{(1)}^{3\\text{\/}2}+\\frac{2}{5}{(1)}^{5\\text{\/}2}\\right]\\hfill \\\\ & =\\frac{256}{15}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Integrating a Function Using the Power Rule.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/v7nDnOyx8Mw?controls=0&amp;start=9&amp;end=96&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\/5.4IntegrationFormulasAndTheNetChangeTheorem9to96_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.4 Integration Formulas and the Net Change Theorem&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572558558\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572150887\">Find the definite integral of [latex]f(x)={x}^{2}-3x[\/latex] over the interval [latex]\\left[1,3\\right].[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572100050\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572100050\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572100050\">[latex]-\\frac{10}{3}[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1170571814551\">Follow the process from <a class=\"autogenerated-content\" href=\"#fs-id1170571609338\">(Figure)<\/a> to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20040\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20040&theme=oea&iframe_resize_id=ohm20040&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1170572627233\" class=\"bc-section section\">\n<h2>The Net Change Theorem<\/h2>\n<p id=\"fs-id1170571583276\">The <strong>net change theorem<\/strong> considers the integral of a <span class=\"no-emphasis\"><em>rate of change<\/em><\/span>. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.<\/p>\n<div id=\"fs-id1170572295126\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Net Change Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1170572480924\">The new value of a changing quantity equals the initial value plus the integral of the rate of change:<\/p>\n<div id=\"fs-id1170572449540\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ F(b)=F(a)+{\\displaystyle\\int }_{a}^{b}F\\text{'}(x)dx\\hfill \\\\ \\hfill \\text{or}\\hfill \\\\ {\\displaystyle\\int }_{a}^{b}F\\text{'}(x)dx=F(b)-F(a).\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1170572110596\">Subtracting [latex]F(a)[\/latex] from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.<\/p>\n<p id=\"fs-id1170572481528\">The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let\u2019s apply the net change theorem to a <span class=\"no-emphasis\">velocity<\/span> function in which the result is <span class=\"no-emphasis\">displacement<\/span>.<\/p>\n<p>We looked at a simple example of this in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/the-definite-integral\/\">The Definite Integral<\/a>. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_05_04_001\">(Figure)<\/a>.<\/p>\n<div style=\"width: 296px\" 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\/11204144\/CNX_Calc_Figure_05_04_002.jpg\" alt=\"A graph with the x axis marked as t and the y axis marked normally. The lines y=40 and y=-30 are drawn over [2,4] and [4,5], respectively.The areas between the lines and the x axis are shaded.\" width=\"286\" height=\"309\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The graph shows speed versus time for the given motion of a car.<\/p>\n<\/div>\n<p id=\"fs-id1170572296931\">Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by<\/p>\n<div id=\"fs-id1170572371074\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{2}^{5}v(t)dt\\hfill & ={\\int }_{2}^{4}40dt+{\\displaystyle\\int }_{4}^{5}-30dt\\hfill \\\\ & =80-30\\hfill \\\\ & =50.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572480929\">Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{2}^{5}|v(t)|dt\\hfill & ={\\int }_{2}^{4}40dt+{\\displaystyle\\int }_{4}^{5}30dt\\hfill \\\\ & =80+30\\hfill \\\\ & =110.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571657334\">Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.<\/p>\n<p id=\"fs-id1170572216807\">To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.<\/p>\n<div id=\"fs-id1170572229863\" class=\"textbox exercises\">\n<h3>Example: Finding Net Displacement<\/h3>\n<p>Given a velocity function [latex]v(t)=3t-5[\/latex] (in meters per second) for a particle in motion from time [latex]t=0[\/latex] to time [latex]t=3,[\/latex] find the net displacement of the particle.<\/p>\n<div id=\"fs-id1170571655282\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572455638\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572455638\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572455638\">Applying the net change theorem, we have<\/p>\n<div id=\"fs-id1170572451466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\int }_{0}^{3}(3t-5)dt\\hfill & =\\frac{3{t}^{2}}{2}-5t{|}_{0}^{3}\\hfill \\\\ \\\\ & =\\left[\\frac{3{(3)}^{2}}{2}-5(3)\\right]-0\\hfill \\\\ & =\\frac{27}{2}-15\\hfill \\\\ & =\\frac{27}{2}-\\frac{30}{2}\\hfill \\\\ & =-\\frac{3}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571655148\">The net displacement is [latex]-\\frac{3}{2}[\/latex] m (<a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_05_04_002\">(Figure)<\/a>).<\/p>\n<div style=\"width: 314px\" 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\/11204148\/CNX_Calc_Figure_05_04_003.jpg\" alt=\"A graph of the line v(t) = 3t \u2013 5, which goes through points (0, -5) and (5\/3, 0). The area over the line and under the x axis in the interval &#091;0, 5\/3&#093; is shaded. The area under the line and above the x axis in the interval &#091;5\/3, 3&#093; is shaded.\" width=\"304\" height=\"422\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The graph shows velocity versus time for a particle moving with a linear velocity function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572247866\" class=\"textbox exercises\">\n<h3>Example: Finding the Total Distance Traveled<\/h3>\n<p>Use <a class=\"autogenerated-content\" href=\"#fs-id1170572229863\">(Figure)<\/a> to find the total distance traveled by a particle according to the velocity function [latex]v(t)=3t-5[\/latex] m\/sec over a time interval [latex]\\left[0,3\\right].[\/latex]<\/p>\n<div id=\"fs-id1170572510360\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572206288\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572206288\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572206288\">The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.<\/p>\n<p id=\"fs-id1170572206422\">To continue with the example, use two integrals to find the total distance. First, find the [latex]t[\/latex]-intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for [latex]t[\/latex].<\/p>\n<p>Thus,<\/p>\n<div id=\"fs-id1170571809306\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}3t-5\\hfill & =\\hfill & 0\\hfill \\\\ \\hfill 3t& =\\hfill & 5\\hfill \\\\ \\hfill t& =\\hfill & \\frac{5}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572522381\">The two subintervals are [latex]\\left[0,\\frac{5}{3}\\right][\/latex] and [latex]\\left[\\frac{5}{3},3\\right].[\/latex] To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval [latex]\\left[0,\\frac{5}{3}\\right],[\/latex] we have [latex]|v(t)|=\\text{\u2212}v(t)[\/latex] over that interval. Over [latex]\\left[\\frac{5}{3},3\\right],[\/latex] the function is positive, so [latex]|v(t)|=v(t).[\/latex] Thus, we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\int }_{0}^{3}|v(t)|dt\\hfill & ={\\int }_{0}^{5\\text{\/}3}\\text{\u2212}v(t)dt+{\\int }_{5\\text{\/}3}^{3}v(t)dt\\hfill \\\\ \\\\ & ={\\int }_{0}^{5\\text{\/}3}5-3tdt+{\\int }_{5\\text{\/}3}^{3}3t-5dt\\hfill \\\\ & ={(5t-\\frac{3{t}^{2}}{2})|}_{0}^{5\\text{\/}3}+{(\\frac{3{t}^{2}}{2}-5t)|}_{5\\text{\/}3}^{3}\\hfill \\\\ & =\\left[5(\\frac{5}{3})-\\frac{3{(5\\text{\/}3)}^{2}}{2}\\right]-0+\\left[\\frac{27}{2}-15\\right]-\\left[\\frac{3{(5\\text{\/}3)}^{2}}{2}-\\frac{25}{3}\\right]\\hfill \\\\ & =\\frac{25}{3}-\\frac{25}{6}+\\frac{27}{2}-15-\\frac{25}{6}+\\frac{25}{3}\\hfill \\\\ & =\\frac{41}{6}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572624276\">So, the total distance traveled is [latex]\\frac{41}{6}[\/latex] m.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Total Distance Traveled.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/v7nDnOyx8Mw?controls=0&amp;start=244&amp;end=456&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\/5.4IntegrationFormulasAndTheNetChangeTheorem244to456_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.4 Integration Formulas and the Net Change Theorem&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572622550\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571609321\">Find the net displacement and total distance traveled in meters given the velocity function [latex]f(t)=\\frac{1}{2}{e}^{t}-2[\/latex] over the interval [latex]\\left[0,2\\right].[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571637281\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571637281\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571637281\">Net displacement: [latex]\\frac{{e}^{2}-9}{2}\\approx -0.8055\\text{m;}[\/latex] total distance traveled: [latex]4\\text{ln}4-7.5+\\frac{{e}^{2}}{2}\\approx 1.740[\/latex] m<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1170572244843\">Follow the procedures from <a class=\"autogenerated-content\" href=\"#fs-id1170572229863\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1170572247866\">(Figure)<\/a>. Note that [latex]f(t)\\le 0[\/latex] for [latex]t\\le \\text{ln}4,[\/latex] and [latex]f(t)\\ge 0[\/latex] for [latex]t\\ge \\text{ln}4.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572547644\" class=\"bc-section section\">\n<h2>Applying the Net Change Theorem<\/h2>\n<p id=\"fs-id1170572294455\">The net change theorem can be applied to the flow and consumption of fluids, as shown in the example below.<\/p>\n<div id=\"fs-id1170571542385\" class=\"textbox exercises\">\n<h3><span style=\"color: #000000;\">Example: How Many Gallons of Gasoline Are Consumed?<\/span><\/h3>\n<p>If the motor on a motorboat is started at [latex]t=0[\/latex] and the boat consumes gasoline at a rate which can be modeled for the first two hours as [latex]5-\\frac{t^{3}}{100}[\/latex] gal\/hr for the first hour, how much gasoline is used in the first hour?<\/p>\n<div id=\"fs-id1170571542387\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572242305\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572242305\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572242305\">Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval [latex]\\left[0,1\\right].[\/latex] We have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int _{0}^{1}}\\left(5-\\dfrac{{t}^{3}}{100}\\right)dt\\hfill & =\\left(5t-\\dfrac{{t}^{4}}{400}\\right){\\displaystyle |_{0}^{1}}\\hfill \\\\ \\\\ \\\\ & =\\left[5(1)-\\dfrac{{(1)}^{4}}{400}\\right]-0\\hfill \\\\ & =5-\\frac{1}{400}\\hfill \\\\ & =4.9975.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572425086\">Thus, the motorboat uses 4.9975 gal of gas in 1 hour.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572379540\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170571609276\">Suppose that, instead of remaining steady during the second half hour of Andrew\u2019s outing, the wind starts to die down according to the function [latex]v(t)=-10t+15.[\/latex] In other words, the wind speed is given by<\/p>\n<p style=\"text-align: center;\">[latex]v(t)=\\bigg\\{\\begin{array}{lll}20t+5\\hfill & \\text{ for }\\hfill & 0\\le t\\le \\frac{1}{2}\\hfill \\\\ -10t+15\\hfill & \\text{ for }\\hfill & \\frac{1}{2}\\le t\\le 1.\\hfill \\end{array}[\/latex]<\/p>\n<p>Under these conditions, how far from his starting point is Andrew after 1 hour?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q39980244\">Hint<\/span><\/p>\n<div id=\"q39980244\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572181954\">Don\u2019t forget that Andrew\u2019s iceboat moves twice as fast as the wind.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571711312\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571711312\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571711312\">17.5 mi<\/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-547\">\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>5.4 Integration Formulas and the Net Change Theorem. <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 1. <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\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/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-1\/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":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at 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