{"id":740,"date":"2025-10-09T15:35:46","date_gmt":"2025-10-09T15:35:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=740"},"modified":"2025-11-20T17:11:00","modified_gmt":"2025-11-20T17:11:00","slug":"vector-calculus-and-vector-fields-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-calculus-and-vector-fields-background-youll-need-2\/","title":{"raw":"Vector Calculus and Vector Fields: Background You'll Need 2","rendered":"Vector Calculus and Vector Fields: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Use the Fundamental Theorem of Calculus to find derivatives of integral functions and calculate definite integrals<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives<\/h2>\r\n<p id=\"fs-id1170571639757\">As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas.<\/p>\r\nThe theorem is comprised of two parts. The first part, which is stated here, establishes the relationship between differentiation and integration.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">Fundamental Theorem of Calculus, Part 1<\/h3>\r\n<p id=\"fs-id1170571679188\">If [latex]f(x)[\/latex] is continuous over an interval [latex]\\left[a,b\\right],[\/latex] and the function [latex]F(x)[\/latex] is defined by<\/p>\r\n\r\n<div id=\"fs-id1170572552076\" class=\"equation\" style=\"text-align: center;\">[latex]F(x)={\\displaystyle\\int }_{a}^{x}f(t)dt,[\/latex]<\/div>\r\n<p id=\"fs-id1170572307182\">then [latex]{F}^{\\prime }(x)=f(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">A couple of points to clarify:\r\n<ol>\r\n \t<li><strong>Notation<\/strong>: We define [latex]F(x) [\/latex] as the definite integral of [latex]f(t)[\/latex] from [latex]a[\/latex] to [latex]x[\/latex]. This might seem confusing because we've said a definite integral is a number. However, [latex]F(x)[\/latex] is a function that gives the value of the definite integral for each [latex]x[\/latex].<\/li>\r\n \t<li><strong>Implications<\/strong>: The Fundamental Theorem of Calculus is crucial because it shows that integration and differentiation are inverse processes. It guarantees that any continuous function [latex]f(x)[\/latex] has an antiderivative [latex]F(x)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1170572498741\">Use the first part of the Fundamental Theorem of Calculus to find the derivative of:<\/p>\r\n\r\n<div id=\"fs-id1170572492494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)={\\displaystyle\\int }_{1}^{x}\\dfrac{1}{{t}^{3}+1}dt.[\/latex]<\/div>\r\n[reveal-answer q=\"fs-id1170572346816\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572346816\"]\r\n<p id=\"fs-id1170572346816\">According to the Fundamental Theorem of Calculus, the derivative is given by:<\/p>\r\n\r\n<div id=\"fs-id1170572346820\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{g}^{\\prime }(x)=\\frac{1}{{x}^{3}+1}.[\/latex][\/hidden-answer]<\/div>\r\n<\/section><section class=\"textbox example\">Let [latex]F(x)={\\displaystyle\\int }_{1}^{\\sqrt{x}} \\sin tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]\r\n<div id=\"fs-id1170572512075\" class=\"exercise\">[reveal-answer q=\"fs-id1170572103655\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572103655\"]\r\n<p id=\"fs-id1170572103655\">Letting [latex]u(x)=\\sqrt{x},[\/latex] we have [latex]F(x)={\\displaystyle\\int }_{1}^{u(x)} \\sin tdt.[\/latex] Thus, by the Fundamental Theorem of Calculus and the chain rule,<\/p>\r\n\r\n<div id=\"fs-id1170572452450\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ {F}^{\\prime }(x)\\hfill &amp; = \\sin (u(x))\\frac{du}{dx}\\hfill \\\\ &amp; = \\sin (u(x))\u00b7(\\frac{1}{2}{x}^{-1\\text{\/}2})\\hfill \\\\ &amp; =\\frac{ \\sin \\sqrt{x}}{2\\sqrt{x}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/section><section class=\"textbox example\">Let [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]\r\n<div id=\"fs-id1170572228880\" class=\"exercise\">[reveal-answer q=\"fs-id1170572142330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572142330\"]\r\n<p id=\"fs-id1170572142330\">We have [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Both limits of integration are variable, so we need to split this into two integrals. We get<\/p>\r\n\r\n<div id=\"fs-id1170572141627\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ F(x)\\hfill &amp; ={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{x}^{0}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\hfill \\\\ &amp; =\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571610445\">Differentiating the first term, we obtain<\/p>\r\n\r\n<div id=\"fs-id1170572245919\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]=\\text{\u2212}{x}^{3}.[\/latex]<\/div>\r\n<p id=\"fs-id1170572305916\">Differentiating the second term, we first let [latex]u(x)=2x.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170572168674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill &amp; =\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{u(x)}{t}^{3}dt\\right]\\hfill \\\\ &amp; ={(u(x))}^{3}\\frac{du}{dx}\\hfill \\\\ &amp; ={(2x)}^{3}\u00b72\\hfill \\\\ &amp; =16{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572207052\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1170572336986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {F}^{\\prime }(x)\\hfill &amp; =\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]+\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill \\\\ &amp; =\\text{\u2212}{x}^{3}+16{x}^{3}\\hfill \\\\ &amp; =15{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=707&amp;end=831&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.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus707to831_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<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]288431[\/ohm_question]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]288432[\/ohm_question]\r\n\r\n<\/section>\r\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\r\nThe Fundamental Theorem of Calculus, Part 2 (also called the <strong>Evaluation Theorem<\/strong>) provides a direct method for evaluating definite integrals. Instead of approximating area with rectangles, you can find an antiderivative and evaluate it at the endpoints.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\r\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then:<\/p>\r\n\r\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\r\n\r\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\r\n[reveal-answer q=\"fs-id1170572173704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572173704\"]\r\n<p id=\"fs-id1170572173704\">Recall the power rule for antiderivatives:<\/p>\r\n\r\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\r\n\r\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill &amp; =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ &amp; =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ &amp; =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ &amp; =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ &amp; =\\frac{16}{3}-16\\hfill \\\\ &amp; =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572135831\"><strong>Analysis<\/strong><\/p>\r\n<p id=\"fs-id1170572135837\">Notice that we did not include the \u201c+ <em>C<\/em>\u201d term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, <em>any<\/em> antiderivative works. So, for convenience, we chose the antiderivative with [latex]C=0.[\/latex] If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.<\/p>\r\n<p id=\"fs-id1170571660070\">The region of the area we just calculated is depicted in Figure 3. Note that the region between the curve and the [latex]x[\/latex]-axis is all below the [latex]x[\/latex]-axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.<\/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\/11204110\/CNX_Calc_Figure_05_03_004.jpg\" alt=\"The graph of the parabola f(t) = t^2 \u2013 4 over [-4, 4]. The area above the curve and below the x axis over [-2, 2] is shaded.\" width=\"325\" height=\"283\" \/> Figure 3. The evaluation of a definite integral can produce a negative value, even though area is always positive.[\/caption]\r\n[\/hidden-answer]<\/section><section class=\"textbox example\">Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex][reveal-answer q=\"fs-id1170571639103\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571639103\"][latex]\\frac{7}{24}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after [latex]5[\/latex] sec wins a prize. If James can skate at a velocity of [latex]f(t)=5+2t[\/latex] ft\/sec and Kathy can skate at a velocity of [latex]g(t)=10+ \\cos (\\frac{\\pi }{2}t)[\/latex] ft\/sec, who is going to win the race?[reveal-answer q=\"fs-id1170572332613\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572332613\"]\r\n<p id=\"fs-id1170572332613\">We need to integrate both functions over the interval [latex]\\left[0,5\\right][\/latex] and see which value is bigger. For James, we want to calculate<\/p>\r\n\r\n<div id=\"fs-id1170572332632\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}(5+2t)dt.[\/latex]<\/div>\r\n<p id=\"fs-id1170571758991\">Using the power rule, we have<\/p>\r\n\r\n<div id=\"fs-id1170571758994\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}(5+2t)dt\\hfill &amp; ={(5t+{t}^{2})|}_{0}^{5}\\hfill \\\\ &amp; =(25+25)=50.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571772586\">Thus, James has skated [latex]50[\/latex] ft after [latex]5[\/latex] sec. Turning now to Kathy, we want to calculate<\/p>\r\n\r\n<div id=\"fs-id1170571772589\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt.[\/latex]<\/div>\r\n<p id=\"fs-id1170572292292\">We know [latex] \\sin t[\/latex] is an antiderivative of [latex] \\cos t,[\/latex] so it is reasonable to expect that an antiderivative of [latex] \\cos (\\frac{\\pi }{2}t)[\/latex] would involve [latex] \\sin (\\frac{\\pi }{2}t).[\/latex] However, when we differentiate [latex] \\sin (\\frac{\\pi }{2}t),[\/latex] we get [latex]\\frac{\\pi }{2} \\cos (\\frac{\\pi }{2}t)[\/latex] as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain<\/p>\r\n\r\n<div id=\"fs-id1170571599308\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt\\hfill &amp; ={(10t+\\frac{2}{\\pi } \\sin (\\frac{\\pi }{2}t))|}_{0}^{5}\\hfill \\\\ &amp; =(50+\\frac{2}{\\pi })-(0-\\frac{2}{\\pi } \\sin 0)\\hfill \\\\ &amp; \\approx 50.6.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572368405\">Kathy has skated approximately [latex]50.6[\/latex] ft after [latex]5[\/latex] sec. Kathy wins, but not by much!<\/p>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=1072&amp;end=1258&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.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus1072to1258_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>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Use the Fundamental Theorem of Calculus to find derivatives of integral functions and calculate definite integrals<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives<\/h2>\n<p id=\"fs-id1170571639757\">As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas.<\/p>\n<p>The theorem is comprised of two parts. The first part, which is stated here, establishes the relationship between differentiation and integration.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">Fundamental Theorem of Calculus, Part 1<\/h3>\n<p id=\"fs-id1170571679188\">If [latex]f(x)[\/latex] is continuous over an interval [latex]\\left[a,b\\right],[\/latex] and the function [latex]F(x)[\/latex] is defined by<\/p>\n<div id=\"fs-id1170572552076\" class=\"equation\" style=\"text-align: center;\">[latex]F(x)={\\displaystyle\\int }_{a}^{x}f(t)dt,[\/latex]<\/div>\n<p id=\"fs-id1170572307182\">then [latex]{F}^{\\prime }(x)=f(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">A couple of points to clarify:<\/p>\n<ol>\n<li><strong>Notation<\/strong>: We define [latex]F(x)[\/latex] as the definite integral of [latex]f(t)[\/latex] from [latex]a[\/latex] to [latex]x[\/latex]. This might seem confusing because we&#8217;ve said a definite integral is a number. However, [latex]F(x)[\/latex] is a function that gives the value of the definite integral for each [latex]x[\/latex].<\/li>\n<li><strong>Implications<\/strong>: The Fundamental Theorem of Calculus is crucial because it shows that integration and differentiation are inverse processes. It guarantees that any continuous function [latex]f(x)[\/latex] has an antiderivative [latex]F(x)[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572498741\">Use the first part of the Fundamental Theorem of Calculus to find the derivative of:<\/p>\n<div id=\"fs-id1170572492494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)={\\displaystyle\\int }_{1}^{x}\\dfrac{1}{{t}^{3}+1}dt.[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572346816\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572346816\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572346816\">According to the Fundamental Theorem of Calculus, the derivative is given by:<\/p>\n<div id=\"fs-id1170572346820\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{g}^{\\prime }(x)=\\frac{1}{{x}^{3}+1}.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Let [latex]F(x)={\\displaystyle\\int }_{1}^{\\sqrt{x}} \\sin tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<div id=\"fs-id1170572512075\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572103655\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572103655\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572103655\">Letting [latex]u(x)=\\sqrt{x},[\/latex] we have [latex]F(x)={\\displaystyle\\int }_{1}^{u(x)} \\sin tdt.[\/latex] Thus, by the Fundamental Theorem of Calculus and the chain rule,<\/p>\n<div id=\"fs-id1170572452450\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ {F}^{\\prime }(x)\\hfill & = \\sin (u(x))\\frac{du}{dx}\\hfill \\\\ & = \\sin (u(x))\u00b7(\\frac{1}{2}{x}^{-1\\text{\/}2})\\hfill \\\\ & =\\frac{ \\sin \\sqrt{x}}{2\\sqrt{x}}.\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Let [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<div id=\"fs-id1170572228880\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572142330\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572142330\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572142330\">We have [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Both limits of integration are variable, so we need to split this into two integrals. We get<\/p>\n<div id=\"fs-id1170572141627\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ F(x)\\hfill & ={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt\\hfill \\\\ & ={\\displaystyle\\int }_{x}^{0}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\hfill \\\\ & =\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571610445\">Differentiating the first term, we obtain<\/p>\n<div id=\"fs-id1170572245919\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]=\\text{\u2212}{x}^{3}.[\/latex]<\/div>\n<p id=\"fs-id1170572305916\">Differentiating the second term, we first let [latex]u(x)=2x.[\/latex] Then,<\/p>\n<div id=\"fs-id1170572168674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill & =\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{u(x)}{t}^{3}dt\\right]\\hfill \\\\ & ={(u(x))}^{3}\\frac{du}{dx}\\hfill \\\\ & ={(2x)}^{3}\u00b72\\hfill \\\\ & =16{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572207052\">Thus,<\/p>\n<div id=\"fs-id1170572336986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {F}^{\\prime }(x)\\hfill & =\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]+\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill \\\\ & =\\text{\u2212}{x}^{3}+16{x}^{3}\\hfill \\\\ & =15{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\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=707&amp;end=831&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.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus707to831_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<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm288431\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288431&theme=lumen&iframe_resize_id=ohm288431&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm288432\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288432&theme=lumen&iframe_resize_id=ohm288432&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\n<p>The Fundamental Theorem of Calculus, Part 2 (also called the <strong>Evaluation Theorem<\/strong>) provides a direct method for evaluating definite integrals. Instead of approximating area with rectangles, you can find an antiderivative and evaluate it at the endpoints.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then:<\/p>\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572173704\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572173704\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572173704\">Recall the power rule for antiderivatives:<\/p>\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill & =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ & =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ & =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ & =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ & =\\frac{16}{3}-16\\hfill \\\\ & =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572135831\"><strong>Analysis<\/strong><\/p>\n<p id=\"fs-id1170572135837\">Notice that we did not include the \u201c+ <em>C<\/em>\u201d term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, <em>any<\/em> antiderivative works. So, for convenience, we chose the antiderivative with [latex]C=0.[\/latex] If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.<\/p>\n<p id=\"fs-id1170571660070\">The region of the area we just calculated is depicted in Figure 3. Note that the region between the curve and the [latex]x[\/latex]-axis is all below the [latex]x[\/latex]-axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.<\/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\/11204110\/CNX_Calc_Figure_05_03_004.jpg\" alt=\"The graph of the parabola f(t) = t^2 \u2013 4 over &#091;-4, 4&#093;. The area above the curve and below the x axis over &#091;-2, 2&#093; is shaded.\" width=\"325\" height=\"283\" \/><figcaption class=\"wp-caption-text\">Figure 3. The evaluation of a definite integral can produce a negative value, even though area is always positive.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571639103\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571639103\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{7}{24}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after [latex]5[\/latex] sec wins a prize. If James can skate at a velocity of [latex]f(t)=5+2t[\/latex] ft\/sec and Kathy can skate at a velocity of [latex]g(t)=10+ \\cos (\\frac{\\pi }{2}t)[\/latex] ft\/sec, who is going to win the race?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572332613\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572332613\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572332613\">We need to integrate both functions over the interval [latex]\\left[0,5\\right][\/latex] and see which value is bigger. For James, we want to calculate<\/p>\n<div id=\"fs-id1170572332632\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}(5+2t)dt.[\/latex]<\/div>\n<p id=\"fs-id1170571758991\">Using the power rule, we have<\/p>\n<div id=\"fs-id1170571758994\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}(5+2t)dt\\hfill & ={(5t+{t}^{2})|}_{0}^{5}\\hfill \\\\ & =(25+25)=50.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571772586\">Thus, James has skated [latex]50[\/latex] ft after [latex]5[\/latex] sec. Turning now to Kathy, we want to calculate<\/p>\n<div id=\"fs-id1170571772589\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt.[\/latex]<\/div>\n<p id=\"fs-id1170572292292\">We know [latex]\\sin t[\/latex] is an antiderivative of [latex]\\cos t,[\/latex] so it is reasonable to expect that an antiderivative of [latex]\\cos (\\frac{\\pi }{2}t)[\/latex] would involve [latex]\\sin (\\frac{\\pi }{2}t).[\/latex] However, when we differentiate [latex]\\sin (\\frac{\\pi }{2}t),[\/latex] we get [latex]\\frac{\\pi }{2} \\cos (\\frac{\\pi }{2}t)[\/latex] as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain<\/p>\n<div id=\"fs-id1170571599308\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt\\hfill & ={(10t+\\frac{2}{\\pi } \\sin (\\frac{\\pi }{2}t))|}_{0}^{5}\\hfill \\\\ & =(50+\\frac{2}{\\pi })-(0-\\frac{2}{\\pi } \\sin 0)\\hfill \\\\ & \\approx 50.6.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572368405\">Kathy has skated approximately [latex]50.6[\/latex] ft after [latex]5[\/latex] sec. Kathy wins, but not by much!<\/p>\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=1072&amp;end=1258&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\/5.3TheFundamentalTheoremOfCalculus1072to1258_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","protected":false},"author":15,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-740","chapter","type-chapter","status-publish","hentry"],"part":436,"module-header":"background_you_need","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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/740","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/740\/revisions"}],"predecessor-version":[{"id":2673,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/740\/revisions\/2673"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/436"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/740\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=740"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=740"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=740"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=740"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}