{"id":391,"date":"2025-02-13T19:44:33","date_gmt":"2025-02-13T19:44:33","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-fundamental-theorem-of-calculus-fresh-take\/"},"modified":"2025-02-13T19:44:33","modified_gmt":"2025-02-13T19:44:33","slug":"the-fundamental-theorem-of-calculus-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-fundamental-theorem-of-calculus-fresh-take\/","title":{"raw":"The Fundamental Theorem of Calculus: Fresh Take","rendered":"The Fundamental Theorem of Calculus: Fresh Take"},"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<h2>The Mean Value Theorem for Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Theorem Statement:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], then there exists at least one point [latex]c \\in [a,b][\/latex] such that: [latex]f(c) = \\frac{1}{b-a} \\int_a^b f(x) dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Geometric Interpretation:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]f(c)[\/latex] equals the average value of the function over [latex][a,b][\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">The area of the rectangle with base [latex]b-a[\/latex] and height [latex]f(c)[\/latex] equals the area under the curve<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Alternative Form:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]\\int_a^b f(x) dx = f(c)(b-a)[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Significance:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Guarantees the existence of a point where the function takes on its average value<\/li>\n\t<li class=\"whitespace-normal break-words\">Bridges discrete and continuous concepts of average<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">The Mean Value Theorem for Integrals provides the theoretical basis for the average value of a function<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\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<p>[reveal-answer q=\"168934\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"168934\"]<\/p>\n<p>Average value [latex]=1.5[\/latex] ; [latex]c=3[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\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\n[reveal-answer q=\"fs-id1170571653986\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571653986\"]\n\n<p id=\"fs-id1170571653986\">[latex]c=\\sqrt{3}[\/latex]<\/p>\n\n[\/hidden-answer]<\/section>\n<h2>Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Theorem Statement:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], and [latex]F(x) = \\int_a^x f(t) dt[\/latex], then [latex]F'(x) = f(x)[\/latex] for all [latex]x \\in [a,b][\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key Implications:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Integration and differentiation are inverse processes<\/li>\n\t<li class=\"whitespace-normal break-words\">Every continuous function has an antiderivative<\/li>\n\t<li class=\"whitespace-normal break-words\">Provides a way to evaluate definite integrals without using Riemann sums<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Interpretation of [latex]F(x)[\/latex]:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]F(x)[\/latex] is a function that gives the area under the curve of [latex]f(t)[\/latex] from [latex]a[\/latex] to [latex]x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">The rate of change of this area function is the original function [latex]f(x)[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Connection to Anti-differentiation:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]F(x)[\/latex] is an antiderivative of [latex]f(x)[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">All antiderivatives of [latex]f(x)[\/latex] differ by a constant<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Applications:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Simplifies the process of finding antiderivatives<\/li>\n\t<li class=\"whitespace-normal break-words\">Allows for the evaluation of complex definite integrals<\/li>\n\t<li class=\"whitespace-normal break-words\">Forms the basis for many advanced calculus techniques<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of [latex]g(r)={\\displaystyle\\int }_{0}^{r}\\sqrt{{x}^{2}+4}dx.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1170571586143\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571586143\"] [latex]{g}^{\\prime }(r)=\\sqrt{{r}^{2}+4}[\/latex] [\/hidden-answer]<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]F(x)={\\displaystyle\\int }_{1}^{{x}^{3}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572557217\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572557217\"]<\/p>\n<p>[latex]{F}^{\\prime }(x)=3{x}^{2} \\cos {x}^{3}[\/latex]<br>\n[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]F(x)={\\displaystyle\\int }_{x}^{{x}^{2}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1170572508003\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572508003\"]\n\n<p id=\"fs-id1170572508003\">[latex]{F}^{\\prime }(x)=2x \\cos {x}^{2}- \\cos x[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Theorem Statement:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">If [latex]f[\/latex] is continuous on [latex][a,b][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x)[\/latex], then: [latex]\\int_a^b f(x)dx = F(b) - F(a)[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key Implications:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Provides a simple method to evaluate definite integrals<\/li>\n\t<li class=\"whitespace-normal break-words\">Connects the concept of area under a curve to the difference of antiderivative values<\/li>\n\t<li class=\"whitespace-normal break-words\">Eliminates the need for limits of Riemann sums in many cases<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Notation:\n\n<ul>\n\t<li class=\"whitespace-normal break-words\">[latex]F(b) - F(a)[\/latex] is often written as [latex]F(x)|_a^b[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Important Points:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Any antiderivative can be used (constant of integration cancels out)<\/li>\n\t<li class=\"whitespace-normal break-words\">The theorem applies to continuous functions on a closed interval<\/li>\n\t<li class=\"whitespace-normal break-words\">The result can be positive, negative, or zero, depending on the function and interval<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Applications:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Simplifies calculations in physics, engineering, and economics<\/li>\n\t<li class=\"whitespace-normal break-words\">Allows for quick evaluation of accumulated change over an interval<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572130389\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572130389\"]<\/p>\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill &amp; ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ &amp; =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ &amp; =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ &amp; =18-6-\\frac{2}{3}+2\\hfill \\\\ &amp; =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572522399\">See Figure 4.<\/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\/11204112\/CNX_Calc_Figure_05_03_005.jpg\" alt=\"The graph of the function f(x) = (x-1) \/ sqrt(x) over [0,9]. The area under the graph over [1,9] is shaded.\" width=\"325\" height=\"246\"> Figure 4. The area under the curve from [latex]x=1[\/latex] to [latex]x=9[\/latex] can be calculated by evaluating a definite integral.[\/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=998&amp;end=1063&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\/5.3TheFundamentalTheoremOfCalculus998to1063_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>.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572368419\">Suppose James and Kathy have a rematch, but this time the official stops the contest after only [latex]3[\/latex] sec. Does this change the outcome?<\/p>\n<p>[reveal-answer q=\"fs-id1170571773427\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571773427\"]<\/p>\n<p id=\"fs-id1170571773427\">Kathy still wins, but by a much larger margin: James skates [latex] 24[\/latex] ft in [latex]3[\/latex] sec, but Kathy skates [latex]29.3634[\/latex] ft in [latex]3[\/latex] sec.<\/p>\n<p>[\/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<h2>The Mean Value Theorem for Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Theorem Statement:\n<ul>\n<li class=\"whitespace-normal break-words\">If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], then there exists at least one point [latex]c \\in [a,b][\/latex] such that: [latex]f(c) = \\frac{1}{b-a} \\int_a^b f(x) dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Geometric Interpretation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]f(c)[\/latex] equals the average value of the function over [latex][a,b][\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">The area of the rectangle with base [latex]b-a[\/latex] and height [latex]f(c)[\/latex] equals the area under the curve<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Alternative Form:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]\\int_a^b f(x) dx = f(c)(b-a)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Significance:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Guarantees the existence of a point where the function takes on its average value<\/li>\n<li class=\"whitespace-normal break-words\">Bridges discrete and continuous concepts of average<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">The Mean Value Theorem for Integrals provides the theoretical basis for the average value of a function<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\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<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q168934\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\">\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 class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571653986\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571653986\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571653986\">[latex]c=\\sqrt{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Theorem Statement:\n<ul>\n<li class=\"whitespace-normal break-words\">If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], and [latex]F(x) = \\int_a^x f(t) dt[\/latex], then [latex]F'(x) = f(x)[\/latex] for all [latex]x \\in [a,b][\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Implications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Integration and differentiation are inverse processes<\/li>\n<li class=\"whitespace-normal break-words\">Every continuous function has an antiderivative<\/li>\n<li class=\"whitespace-normal break-words\">Provides a way to evaluate definite integrals without using Riemann sums<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Interpretation of [latex]F(x)[\/latex]:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]F(x)[\/latex] is a function that gives the area under the curve of [latex]f(t)[\/latex] from [latex]a[\/latex] to [latex]x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">The rate of change of this area function is the original function [latex]f(x)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Connection to Anti-differentiation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]F(x)[\/latex] is an antiderivative of [latex]f(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">All antiderivatives of [latex]f(x)[\/latex] differ by a constant<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Applications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Simplifies the process of finding antiderivatives<\/li>\n<li class=\"whitespace-normal break-words\">Allows for the evaluation of complex definite integrals<\/li>\n<li class=\"whitespace-normal break-words\">Forms the basis for many advanced calculus techniques<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of [latex]g(r)={\\displaystyle\\int }_{0}^{r}\\sqrt{{x}^{2}+4}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571586143\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571586143\" class=\"hidden-answer\" style=\"display: none\"> [latex]{g}^{\\prime }(r)=\\sqrt{{r}^{2}+4}[\/latex] <\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]F(x)={\\displaystyle\\int }_{1}^{{x}^{3}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572557217\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572557217\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{F}^{\\prime }(x)=3{x}^{2} \\cos {x}^{3}[\/latex]\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]F(x)={\\displaystyle\\int }_{x}^{{x}^{2}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572508003\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572508003\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572508003\">[latex]{F}^{\\prime }(x)=2x \\cos {x}^{2}- \\cos x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Theorem Statement:\n<ul>\n<li class=\"whitespace-normal break-words\">If [latex]f[\/latex] is continuous on [latex][a,b][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x)[\/latex], then: [latex]\\int_a^b f(x)dx = F(b) - F(a)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Implications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Provides a simple method to evaluate definite integrals<\/li>\n<li class=\"whitespace-normal break-words\">Connects the concept of area under a curve to the difference of antiderivative values<\/li>\n<li class=\"whitespace-normal break-words\">Eliminates the need for limits of Riemann sums in many cases<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Notation:\n<ul>\n<li class=\"whitespace-normal break-words\">[latex]F(b) - F(a)[\/latex] is often written as [latex]F(x)|_a^b[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Important Points:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Any antiderivative can be used (constant of integration cancels out)<\/li>\n<li class=\"whitespace-normal break-words\">The theorem applies to continuous functions on a closed interval<\/li>\n<li class=\"whitespace-normal break-words\">The result can be positive, negative, or zero, depending on the function and interval<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Applications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Simplifies calculations in physics, engineering, and economics<\/li>\n<li class=\"whitespace-normal break-words\">Allows for quick evaluation of accumulated change over an interval<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572130389\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572130389\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill & ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ & =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ & =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ & =18-6-\\frac{2}{3}+2\\hfill \\\\ & =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572522399\">See Figure 4.<\/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\/11204112\/CNX_Calc_Figure_05_03_005.jpg\" alt=\"The graph of the function f(x) = (x-1) \/ sqrt(x) over &#091;0,9&#093;. The area under the graph over &#091;1,9&#093; is shaded.\" width=\"325\" height=\"246\" \/><figcaption class=\"wp-caption-text\">Figure 4. The area under the curve from [latex]x=1[\/latex] to [latex]x=9[\/latex] can be calculated by evaluating a definite integral.<\/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=998&amp;end=1063&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\/5.3TheFundamentalTheoremOfCalculus998to1063_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>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572368419\">Suppose James and Kathy have a rematch, but this time the official stops the contest after only [latex]3[\/latex] sec. Does this change the outcome?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571773427\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571773427\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571773427\">Kathy still wins, but by a much larger margin: James skates [latex]24[\/latex] ft in [latex]3[\/latex] sec, but Kathy skates [latex]29.3634[\/latex] ft in [latex]3[\/latex] sec.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":20,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":371,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/391"}],"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\/391\/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\/391\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=391"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=391"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=391"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=391"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}