{"id":545,"date":"2021-02-04T16:10:10","date_gmt":"2021-02-04T16:10:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=545"},"modified":"2022-03-16T21:29:29","modified_gmt":"2022-03-16T21:29:29","slug":"the-mean-value-theorem-for-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/the-mean-value-theorem-for-integrals\/","title":{"raw":"The Mean Value Theorem for Integrals","rendered":"The Mean Value Theorem for Integrals"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the meaning of the Mean Value Theorem for Integrals<\/li>\r\n<\/ul>\r\n<\/div>\r\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] 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>\r\n\r\n<div id=\"fs-id1170572224851\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Mean Value Theorem for Integrals<\/h3>\r\n\r\n<hr \/>\r\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>\r\n\r\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>\r\n&nbsp;\r\n<p id=\"fs-id1170572370703\">This formula can also be stated as<\/p>\r\n\r\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>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572087581\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1170572169022\">Since [latex]f(x)[\/latex] is continuous on [latex]\\left[a,b\\right],[\/latex] by the extreme value theorem (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/maxima-and-minima\/\">Maxima and Minima<\/a>), 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 (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/the-definite-integral\/\">The Definite Integral<\/a>), we have<\/p>\r\n\r\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>\r\n&nbsp;\r\n<p id=\"fs-id1170572622216\">Dividing by [latex]b-a[\/latex] gives us<\/p>\r\n\r\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>\r\n&nbsp;\r\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 (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/continuity\/\">Continuity<\/a>), there is a number [latex]c[\/latex] over [latex]\\left[a,b\\right][\/latex] such that<\/p>\r\n\r\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>\r\n&nbsp;\r\n<p id=\"fs-id1170572224806\">and the proof is complete.<\/p>\r\n<p id=\"fs-id1170572421834\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170572141909\" class=\"textbook exercises\">\r\n<div id=\"fs-id1170572306320\" class=\"exercise\">\r\n<h3>Example: Finding the Average Value of a Function<\/h3>\r\nFind 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]\r\n[reveal-answer q=\"fs-id1170572114676\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572114676\"]The formula states the mean value of [latex]f(x)[\/latex] is given by\r\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>\r\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>\r\n\r\n<div id=\"fs-id1170572375674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\frac{1}{2}(4)(8)=16.[\/latex]<\/div>\r\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>\r\n\r\n<div id=\"fs-id1170572094543\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{4}(16)=4.[\/latex]<\/div>\r\n<p id=\"fs-id1170572559549\">Set the average value equal to [latex]f(c)[\/latex] and solve for [latex]c[\/latex].<\/p>\r\n\r\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>\r\n<p id=\"fs-id1170572111868\">At [latex]c=2,f(2)=4.[\/latex]<\/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\/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][\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding the Average Value of a Function.\r\n\r\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\"><\/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\/5.3TheFundamentalTheoremOfCalculus76to215_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]\r\n<div id=\"fs-id1170572332976\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the average value of the function [latex]f(x)=\\dfrac{x}{2}[\/latex] over the interval [latex]\\left[0,6\\right][\/latex] and find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of the function over [latex]\\left[0,6\\right].[\/latex]\r\n\r\n[reveal-answer q=\"168934\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"168934\"]\r\n\r\nAverage value [latex]=1.5[\/latex] ; [latex]c=3[\/latex]\r\n\r\n[\/hidden-answer]\r\n<h4>Hint<\/h4>\r\nUse the procedures from the last example to solve the problem\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572451873\" class=\"textbook exercises\">\r\n<div id=\"fs-id1170571654400\" class=\"exercise\">\r\n<h3>example: FINDING THE POINT WHERE A FUNCTION TAKES ON ITS AVERAGE VALUE<\/h3>\r\nGiven [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]\r\n[reveal-answer q=\"fs-id1170571816124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571816124\"]\r\n<p id=\"fs-id1170571816124\">We are looking for the value of [latex]c[\/latex] such that<\/p>\r\n\r\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>\r\n<p id=\"fs-id1170572135349\">Replacing [latex]f(c)[\/latex] with [latex]c[\/latex]<sup>2<\/sup>, we have<\/p>\r\n\r\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>\r\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>\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\/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][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div>\r\n\r\nWatch the following video to see the worked solution to Example: Finding the Point Where a Function Takes on its Average Value.\r\n\r\n<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\"><\/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\/5.3TheFundamentalTheoremOfCalculus218to299_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]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571569107\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]{\\displaystyle\\int }_{0}^{3}(2{x}^{2}-1)dx=15,[\/latex] find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of [latex]f(x)=2{x}^{2}-1[\/latex] over [latex]\\left[0,3\\right].[\/latex]\r\n<div id=\"fs-id1170572176911\" class=\"exercise\">[reveal-answer q=\"fs-id1170571653986\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571653986\"]\r\n<p id=\"fs-id1170571653986\">[latex]c=\\sqrt{3}[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170572137926\">Use the procedures from the last example to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h2><\/h2>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the meaning of the Mean Value Theorem for Integrals<\/li>\n<\/ul>\n<\/div>\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] 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<div id=\"fs-id1170572224851\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Mean Value Theorem for Integrals<\/h3>\n<hr \/>\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>&nbsp;<\/p>\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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1170572087581\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1170572169022\">Since [latex]f(x)[\/latex] is continuous on [latex]\\left[a,b\\right],[\/latex] by the extreme value theorem (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/maxima-and-minima\/\">Maxima and Minima<\/a>), 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 (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/the-definite-integral\/\">The Definite Integral<\/a>), 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>&nbsp;<\/p>\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>&nbsp;<\/p>\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 (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/continuity\/\">Continuity<\/a>), 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>&nbsp;<\/p>\n<p id=\"fs-id1170572224806\">and the proof is complete.<\/p>\n<p id=\"fs-id1170572421834\">[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1170572141909\" class=\"textbook exercises\">\n<div id=\"fs-id1170572306320\" class=\"exercise\">\n<h3>Example: Finding the Average Value of a Function<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572114676\">Show Solution<\/span><\/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<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\/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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Average Value of a Function.<\/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\"><\/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\/5.3TheFundamentalTheoremOfCalculus76to215_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<div id=\"fs-id1170572332976\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the average value of the function [latex]f(x)=\\dfrac{x}{2}[\/latex] over the interval [latex]\\left[0,6\\right][\/latex] and find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of the function over [latex]\\left[0,6\\right].[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q168934\">Show Solution<\/span><\/p>\n<div id=\"q168934\" class=\"hidden-answer\" style=\"display: none\">\n<p>Average value [latex]=1.5[\/latex] ; [latex]c=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4>Hint<\/h4>\n<p>Use the procedures from the last example to solve the problem<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572451873\" class=\"textbook exercises\">\n<div id=\"fs-id1170571654400\" class=\"exercise\">\n<h3>example: FINDING THE POINT WHERE A FUNCTION TAKES ON ITS AVERAGE VALUE<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571816124\">Show Solution<\/span><\/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[\/latex]<sup>2<\/sup>, 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<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\/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\" \/><\/p>\n<p 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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p>Watch the following video to see the worked solution to Example: Finding the Point Where a Function Takes on its Average Value.<\/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\"><\/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\/5.3TheFundamentalTheoremOfCalculus218to299_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<\/div>\n<div id=\"fs-id1170571569107\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given [latex]{\\displaystyle\\int }_{0}^{3}(2{x}^{2}-1)dx=15,[\/latex] find [latex]c[\/latex] such that [latex]f(c)[\/latex] equals the average value of [latex]f(x)=2{x}^{2}-1[\/latex] over [latex]\\left[0,3\\right].[\/latex]<\/p>\n<div id=\"fs-id1170572176911\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571653986\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571653986\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571653986\">[latex]c=\\sqrt{3}[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1170572137926\">Use the procedures from the last example to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2><\/h2>\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-545\">\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>5.3 The Fundamental Theorem of Calculus. <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":14,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-545","chapter","type-chapter","status-publish","hentry"],"part":57,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/545","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":14,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/545\/revisions"}],"predecessor-version":[{"id":4875,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/545\/revisions\/4875"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/57"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/545\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=545"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=545"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=545"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=545"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}