{"id":387,"date":"2025-02-13T19:44:31","date_gmt":"2025-02-13T19:44:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-fundamental-theorem-of-calculus-learn-it-1\/"},"modified":"2025-02-13T19:44:31","modified_gmt":"2025-02-13T19:44:31","slug":"the-fundamental-theorem-of-calculus-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-fundamental-theorem-of-calculus-learn-it-1\/","title":{"raw":"The Fundamental Theorem of Calculus: Learn It 1","rendered":"The Fundamental Theorem of Calculus: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Understand the Mean Value Theorem for Integrals and both components of the Fundamental Theorem of Calculus<\/li>\n\t<li>Use the Fundamental Theorem of Calculus to find derivatives of integral functions and calculate definite integrals<\/li>\n\t<li>Describe how differentiation and integration are interconnected<\/li>\n<\/ul>\n<\/section>\n<p>In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function. Unfortunately, the only tools we have available to calculate the value of a definite integral are geometric area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section, we will look at more powerful and useful techniques for evaluating definite integrals.<\/p>\n<p>These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and explored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s, and it is codified in what we now call the Fundamental Theorem of Calculus.<\/p>\n<section class=\"textbox interact\">\n<p>Isaac <span class=\"no-emphasis\">Newton<\/span>\u2019s contributions to mathematics and physics changed the way we look at the world. The relationships he discovered, codified as Newton\u2019s laws and the law of universal gravitation, are still taught as foundational material in physics today, and his calculus has spawned entire fields of mathematics. <a href=\"https:\/\/www.biography.com\/scientist\/isaac-newton#synopsis\" target=\"_blank\" rel=\"noopener\">To learn more, read a brief biography of Newton with multimedia clips.<\/a><\/p>\n<\/section>\n<h2>The Mean Value Theorem for Integrals<\/h2>\n<p id=\"fs-id1170572295468\">The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if [latex]f(x)[\/latex] is continuous, a point [latex]c[\/latex] exists in an interval [latex]\\left[a,b\\right][\/latex] such that the value of the function at [latex]c[\/latex] is equal to the average value of [latex]f(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\n<p>We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">the mean value theorem for integrals<\/h3>\n<p id=\"fs-id1170572382252\">If [latex]f(x)[\/latex] is continuous over an interval [latex]\\left[a,b\\right],[\/latex] then there is at least one point [latex]c\\in \\left[a,b\\right][\/latex] such that<\/p>\n<div id=\"fs-id1170571654721\" class=\"equation\" style=\"text-align: center;\">[latex]f(c)=\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(x)dx.[\/latex]<\/div>\n<p id=\"fs-id1170572370703\">This formula can also be stated as<\/p>\n<div id=\"fs-id1170572110045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=f(c)(b-a).[\/latex]<\/div>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr>\n<p id=\"fs-id1170572169022\">Since [latex]f(x)[\/latex] is continuous on [latex]\\left[a,b\\right],[\/latex] by the extreme value theorem, it assumes minimum and maximum values\u2014[latex]m[\/latex] and <em>M<\/em>, respectively\u2014on [latex]\\left[a,b\\right].[\/latex] Then, for all [latex]x[\/latex] in [latex]\\left[a,b\\right],[\/latex] we have [latex]m\\le f(x)\\le M.[\/latex] Therefore, by the comparison theorem, we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m(b-a)\\le {\\displaystyle\\int }_{a}^{b}f(x)dx\\le M(b-a).[\/latex]<\/div>\n<p id=\"fs-id1170572622216\">Dividing by [latex]b-a[\/latex] gives us<\/p>\n<div id=\"fs-id1170572228715\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m\\le \\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(x)dx\\le M.[\/latex]<\/div>\n<p id=\"fs-id1170572204800\">Since [latex]\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(x)dx[\/latex] is a number between [latex]m[\/latex] and <em>M<\/em>, and since [latex]f(x)[\/latex] is continuous and assumes the values [latex]m[\/latex] and <em>M<\/em> over [latex]\\left[a,b\\right],[\/latex] by the Intermediate Value Theorem, there is a number [latex]c[\/latex] over [latex]\\left[a,b\\right][\/latex] such that<\/p>\n<div id=\"fs-id1170572096606\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(c)=\\dfrac{1}{b-a}{\\displaystyle\\int}_{a}^{b}f(x)dx,[\/latex]<\/div>\n<p id=\"fs-id1170572224806\">and the proof is complete.<\/p>\n<p id=\"fs-id1170572421834\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the average value of the function [latex]f(x)=8-2x[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] and find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of the function over [latex]\\left[0,4\\right].[\/latex]<br>\n[reveal-answer q=\"fs-id1170572114676\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572114676\"]The formula states the mean value of [latex]f(x)[\/latex] is given by<\/p>\n<div id=\"fs-id1170571553890\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{4-0}{\\displaystyle\\int }_{0}^{4}(8-2x)dx.[\/latex]<\/div>\n<p id=\"fs-id1170572589327\">We can see in Figure 1 that the function represents a straight line and forms a right triangle bounded by the [latex]x[\/latex]- and [latex]y[\/latex]-axes. The area of the triangle is [latex]A=\\frac{1}{2}(\\text{base})(\\text{height}).[\/latex] We have<\/p>\n<div id=\"fs-id1170572375674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\frac{1}{2}(4)(8)=16.[\/latex]<\/div>\n<p id=\"fs-id1170572549185\">The average value is found by multiplying the area by [latex]\\frac{1}{(4-0)}.[\/latex] Thus, the average value of the function is<\/p>\n<div id=\"fs-id1170572094543\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{4}(16)=4.[\/latex]<\/div>\n<p id=\"fs-id1170572559549\">Set the average value equal to [latex]f(c)[\/latex] and solve for [latex]c[\/latex].<\/p>\n<div id=\"fs-id1170572558241\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}8-2c\\hfill &amp; =\\hfill &amp; 4\\hfill \\\\ \\hfill c&amp; =\\hfill &amp; 2\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572111868\">At [latex]c=2,f(2)=4.[\/latex]<\/p>\n\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\/11204104\/CNX_Calc_Figure_05_03_002.jpg\" alt=\"The graph of a decreasing line f(x) = 8 \u2013 2x over [-1,4.5]. The line y=4 is drawn over [0,4], which intersects with the line at (2,4). A line is drawn down from (2,4) to the x axis and from (4,4) to the y axis. The area under y=4 is shaded.\" width=\"325\" height=\"433\"> Figure 1. By the Mean Value Theorem, the continuous function [latex]f(x)[\/latex] takes on its average value at c at least once over a closed interval.[\/caption]\n\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\/UdsTNaiWmbs?controls=0&amp;start=76&amp;end=215&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>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.\n\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus76to215_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3 The Fundamental Theorem of Calculus\" here (opens in new window)<\/a>. [\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Given [latex]{\\displaystyle\\int }_{0}^{3}{x}^{2}dx=9,[\/latex] find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of [latex]f(x)={x}^{2}[\/latex] over [latex]\\left[0,3\\right].[\/latex]<br>\n[reveal-answer q=\"fs-id1170571816124\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571816124\"]<\/p>\n<p id=\"fs-id1170571816124\">We are looking for the value of [latex]c[\/latex] such that<\/p>\n<div id=\"fs-id1170572480974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(c)=\\frac{1}{3-0}{\\displaystyle\\int }_{0}^{3}{x}^{2}dx=\\frac{1}{3}(9)=3.[\/latex]<\/div>\n<p id=\"fs-id1170572135349\">Replacing [latex]f(c)[\/latex] with [latex]c^2[\/latex], we have<\/p>\n<div id=\"fs-id1170572167258\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}{c}^{2}\\hfill &amp; =\\hfill &amp; 3\\hfill \\\\ c\\hfill &amp; =\\hfill &amp; \\text{\u00b1}\\sqrt{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572222654\">Since [latex]\\text{\u2212}\\sqrt{3}[\/latex] is outside the interval, take only the positive value. Thus, [latex]c=\\sqrt{3}[\/latex] (Figure 2).<\/p>\n\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\/11204107\/CNX_Calc_Figure_05_03_003.jpg\" alt=\"A graph of the parabola f(x) = x^2 over [-2, 3]. The area under the curve and above the x axis is shaded, and the point (sqrt(3), 3) is marked.\" width=\"325\" height=\"471\"> Figure 2. Over the interval [latex]\\left[0,3\\right],[\/latex] the function [latex]f(x)={x}^{2}[\/latex] takes on its average value at [latex]c=\\sqrt{3}.[\/latex][\/caption]\nWatch the following video to see the worked solution to this example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=218&amp;end=299&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\/5.3TheFundamentalTheoremOfCalculus218to299_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3 The Fundamental Theorem of Calculus\" here (opens in new window)<\/a>. [\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand the Mean Value Theorem for Integrals and both components of the Fundamental Theorem of Calculus<\/li>\n<li>Use the Fundamental Theorem of Calculus to find derivatives of integral functions and calculate definite integrals<\/li>\n<li>Describe how differentiation and integration are interconnected<\/li>\n<\/ul>\n<\/section>\n<p>In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function. Unfortunately, the only tools we have available to calculate the value of a definite integral are geometric area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section, we will look at more powerful and useful techniques for evaluating definite integrals.<\/p>\n<p>These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and explored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s, and it is codified in what we now call the Fundamental Theorem of Calculus.<\/p>\n<section class=\"textbox interact\">\n<p>Isaac <span class=\"no-emphasis\">Newton<\/span>\u2019s contributions to mathematics and physics changed the way we look at the world. The relationships he discovered, codified as Newton\u2019s laws and the law of universal gravitation, are still taught as foundational material in physics today, and his calculus has spawned entire fields of mathematics. <a href=\"https:\/\/www.biography.com\/scientist\/isaac-newton#synopsis\" target=\"_blank\" rel=\"noopener\">To learn more, read a brief biography of Newton with multimedia clips.<\/a><\/p>\n<\/section>\n<h2>The Mean Value Theorem for Integrals<\/h2>\n<p id=\"fs-id1170572295468\">The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if [latex]f(x)[\/latex] is continuous, a point [latex]c[\/latex] exists in an interval [latex]\\left[a,b\\right][\/latex] such that the value of the function at [latex]c[\/latex] is equal to the average value of [latex]f(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\n<p>We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">the mean value theorem for integrals<\/h3>\n<p id=\"fs-id1170572382252\">If [latex]f(x)[\/latex] is continuous over an interval [latex]\\left[a,b\\right],[\/latex] then there is at least one point [latex]c\\in \\left[a,b\\right][\/latex] such that<\/p>\n<div id=\"fs-id1170571654721\" class=\"equation\" style=\"text-align: center;\">[latex]f(c)=\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(x)dx.[\/latex]<\/div>\n<p id=\"fs-id1170572370703\">This formula can also be stated as<\/p>\n<div id=\"fs-id1170572110045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=f(c)(b-a).[\/latex]<\/div>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1170572169022\">Since [latex]f(x)[\/latex] is continuous on [latex]\\left[a,b\\right],[\/latex] by the extreme value theorem, it assumes minimum and maximum values\u2014[latex]m[\/latex] and <em>M<\/em>, respectively\u2014on [latex]\\left[a,b\\right].[\/latex] Then, for all [latex]x[\/latex] in [latex]\\left[a,b\\right],[\/latex] we have [latex]m\\le f(x)\\le M.[\/latex] Therefore, by the comparison theorem, we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m(b-a)\\le {\\displaystyle\\int }_{a}^{b}f(x)dx\\le M(b-a).[\/latex]<\/div>\n<p id=\"fs-id1170572622216\">Dividing by [latex]b-a[\/latex] gives us<\/p>\n<div id=\"fs-id1170572228715\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m\\le \\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(x)dx\\le M.[\/latex]<\/div>\n<p id=\"fs-id1170572204800\">Since [latex]\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(x)dx[\/latex] is a number between [latex]m[\/latex] and <em>M<\/em>, and since [latex]f(x)[\/latex] is continuous and assumes the values [latex]m[\/latex] and <em>M<\/em> over [latex]\\left[a,b\\right],[\/latex] by the Intermediate Value Theorem, there is a number [latex]c[\/latex] over [latex]\\left[a,b\\right][\/latex] such that<\/p>\n<div id=\"fs-id1170572096606\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(c)=\\dfrac{1}{b-a}{\\displaystyle\\int}_{a}^{b}f(x)dx,[\/latex]<\/div>\n<p id=\"fs-id1170572224806\">and the proof is complete.<\/p>\n<p id=\"fs-id1170572421834\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the average value of the function [latex]f(x)=8-2x[\/latex] over the interval [latex]\\left[0,4\\right][\/latex] and find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of the function over [latex]\\left[0,4\\right].[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572114676\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572114676\" class=\"hidden-answer\" style=\"display: none\">The formula states the mean value of [latex]f(x)[\/latex] is given by<\/p>\n<div id=\"fs-id1170571553890\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{4-0}{\\displaystyle\\int }_{0}^{4}(8-2x)dx.[\/latex]<\/div>\n<p id=\"fs-id1170572589327\">We can see in Figure 1 that the function represents a straight line and forms a right triangle bounded by the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-axes. The area of the triangle is [latex]A=\\frac{1}{2}(\\text{base})(\\text{height}).[\/latex] We have<\/p>\n<div id=\"fs-id1170572375674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\frac{1}{2}(4)(8)=16.[\/latex]<\/div>\n<p id=\"fs-id1170572549185\">The average value is found by multiplying the area by [latex]\\frac{1}{(4-0)}.[\/latex] Thus, the average value of the function is<\/p>\n<div id=\"fs-id1170572094543\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{4}(16)=4.[\/latex]<\/div>\n<p id=\"fs-id1170572559549\">Set the average value equal to [latex]f(c)[\/latex] and solve for [latex]c[\/latex].<\/p>\n<div id=\"fs-id1170572558241\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}8-2c\\hfill & =\\hfill & 4\\hfill \\\\ \\hfill c& =\\hfill & 2\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572111868\">At [latex]c=2,f(2)=4.[\/latex]<\/p>\n<figure style=\"width: 325px\" 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\/11204104\/CNX_Calc_Figure_05_03_002.jpg\" alt=\"The graph of a decreasing line f(x) = 8 \u2013 2x over &#091;-1,4.5&#093;. The line y=4 is drawn over &#091;0,4&#093;, which intersects with the line at (2,4). A line is drawn down from (2,4) to the x axis and from (4,4) to the y axis. The area under y=4 is shaded.\" width=\"325\" height=\"433\" \/><figcaption class=\"wp-caption-text\">Figure 1. By the Mean Value Theorem, the continuous function [latex]f(x)[\/latex] takes on its average value at c at least once over a closed interval.<\/figcaption><\/figure>\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\/UdsTNaiWmbs?controls=0&amp;start=76&amp;end=215&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\/5.3TheFundamentalTheoremOfCalculus76to215_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3 The Fundamental Theorem of Calculus&#8221; here (opens in new window)<\/a>. <\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Given [latex]{\\displaystyle\\int }_{0}^{3}{x}^{2}dx=9,[\/latex] find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of [latex]f(x)={x}^{2}[\/latex] over [latex]\\left[0,3\\right].[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571816124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571816124\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571816124\">We are looking for the value of [latex]c[\/latex] such that<\/p>\n<div id=\"fs-id1170572480974\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(c)=\\frac{1}{3-0}{\\displaystyle\\int }_{0}^{3}{x}^{2}dx=\\frac{1}{3}(9)=3.[\/latex]<\/div>\n<p id=\"fs-id1170572135349\">Replacing [latex]f(c)[\/latex] with [latex]c^2[\/latex], we have<\/p>\n<div id=\"fs-id1170572167258\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}{c}^{2}\\hfill & =\\hfill & 3\\hfill \\\\ c\\hfill & =\\hfill & \\text{\u00b1}\\sqrt{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572222654\">Since [latex]\\text{\u2212}\\sqrt{3}[\/latex] is outside the interval, take only the positive value. Thus, [latex]c=\\sqrt{3}[\/latex] (Figure 2).<\/p>\n<figure style=\"width: 325px\" 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\/11204107\/CNX_Calc_Figure_05_03_003.jpg\" alt=\"A graph of the parabola f(x) = x^2 over &#091;-2, 3&#093;. The area under the curve and above the x axis is shaded, and the point (sqrt(3), 3) is marked.\" width=\"325\" height=\"471\" \/><figcaption class=\"wp-caption-text\">Figure 2. Over the interval [latex]\\left[0,3\\right],[\/latex] the function [latex]f(x)={x}^{2}[\/latex] takes on its average value at [latex]c=\\sqrt{3}.[\/latex]<\/figcaption><\/figure>\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\/UdsTNaiWmbs?controls=0&amp;start=218&amp;end=299&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\/5.3TheFundamentalTheoremOfCalculus218to299_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3 The Fundamental Theorem of Calculus&#8221; here (opens in new window)<\/a>. <\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":16,"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 https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.3 The Fundamental Theorem of Calculus\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":371,"module-header":"","content_attributions":[{"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 https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"5.3 The Fundamental Theorem of Calculus","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/387"}],"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\/387\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/371"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/387\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=387"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=387"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=387"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}