{"id":673,"date":"2025-10-09T15:29:48","date_gmt":"2025-10-09T15:29:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=673"},"modified":"2025-11-20T16:31:53","modified_gmt":"2025-11-20T16:31:53","slug":"double-integration-and-applications-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integration-and-applications-background-youll-need-1\/","title":{"raw":"Double Integration and Applications: Background You'll Need 1","rendered":"Double Integration and Applications: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Evaluate definite integrals using integration techniques including substitution, and integration by parts<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Substitution for Definite Integrals<\/h2>\r\n<p id=\"fs-id1170573387843\">Substitution can be used with definite integrals. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>substitution with definite integrals<\/h3>\r\n<p id=\"fs-id1170570999523\">Let [latex]u=g(x)[\/latex] and let [latex]{g}^{\\text{\u2032}}[\/latex] be continuous over an interval [latex]\\left[a,b\\right],[\/latex] and let [latex]f[\/latex] be continuous over the range of [latex]u=g(x).[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170571189407\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(g(x)){g}^{\\prime }(x)dx={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du[\/latex]<\/div>\r\n<\/div>\r\n<\/section><section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[\/latex][reveal-answer q=\"fs-id1170572587723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587723\"]\r\n<p id=\"fs-id1170573440252\">Let [latex]u=1+2{x}^{3},[\/latex] so [latex]du=6{x}^{2}dx.[\/latex]<\/p>\r\nSince the original function includes one factor of [latex]x^2[\/latex]\u00a0and [latex]du=6{x}^{2}dx,[\/latex] multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{6}.[\/latex]\r\n\r\nThen,\r\n<div id=\"fs-id1170573362498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}du\\hfill &amp; =\\hfill &amp; 6{x}^{2}dx\\hfill \\\\ \\frac{1}{6}du\\hfill &amp; =\\hfill &amp; {x}^{2}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571157085\">To adjust the limits of integration, note that when [latex]x=0,u=1+2(0)=1,[\/latex] and when [latex]x=1,u=1+2(1)=3.[\/latex]<\/p>\r\nThen,\r\n<div id=\"fs-id1170571303367\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du.[\/latex]<\/div>\r\n<p id=\"fs-id1170571285285\">Evaluating this expression, we get<\/p>\r\n\r\n<div id=\"fs-id1170571285289\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du\\hfill &amp; =(\\frac{1}{6})(\\frac{{u}^{6}}{6}){|}_{1}^{3}\\hfill \\\\ &amp; =\\frac{1}{36}\\left[{(3)}^{6}-{(1)}^{6}\\right]\\hfill \\\\ &amp; =\\frac{182}{9}.\\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\/Ak_y3lsBNfE?controls=0&amp;start=519&amp;end=664&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\/5.5Substitution519to664_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.5 Substitution\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.[\/latex][reveal-answer q=\"fs-id1170572587724\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587724\"]\r\n<p id=\"fs-id1170571347206\">Let [latex]u=4{x}^{3}+3.[\/latex] Then, [latex]du=8xdx.[\/latex]<\/p>\r\nTo adjust the limits of integration, we note that when [latex]x=0,u=3,[\/latex] and when [latex]x=1,u=7.[\/latex]\r\n\r\nSo our substitution gives,\r\n<div id=\"fs-id1170571052992\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\\hfill &amp; =\\frac{1}{8}{\\displaystyle\\int }_{3}^{7}{e}^{u}du\\hfill \\\\ \\\\ &amp; =\\frac{1}{8}{e}^{u}{|}_{3}^{7}\\hfill \\\\ &amp; =\\frac{{e}^{7}-{e}^{3}}{8}\\hfill \\\\ &amp; \\approx 134.568.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]211747[\/ohm_question]<\/section><section>\r\n<h2 data-type=\"title\">Integration by Parts for Definite Integrals<\/h2>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<div>\r\n<h3>integration by parts for definite integrals<\/h3>\r\n<p id=\"fs-id1165042231049\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives on [latex]\\left[a,b\\right][\/latex]. Then<\/p>\r\n\r\n<center>[latex]{\\displaystyle\\int }_{a}^{b}udv={uv|}_{a}^{b}-{\\displaystyle\\int }_{a}^{b}vdu[\/latex].<\/center><\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042058875\" data-type=\"problem\">\u00a0Find the area of the region bounded above by the graph of [latex]y={\\tan}^{-1}x[\/latex] and below by the [latex]x[\/latex] -axis over the interval [latex]\\left[0,1\\right][\/latex].<\/div>\r\n<div id=\"fs-id1165042071427\" data-type=\"commentary\">\r\n\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1165042237897\" data-type=\"solution\">\r\n<p id=\"fs-id1165042237899\">This region is shown in Figure 1. To find the area, we must evaluate [latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}{\\tan}^{-1}xdx[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_01_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"562\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233750\/CNX_Calc_Figure_07_01_001.jpg\" alt=\"This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.\" width=\"562\" height=\"570\" data-media-type=\"image\/jpeg\" \/> Figure 1. To find the area of the shaded region, we have to use integration by parts.[\/caption]<\/figure>\r\n<p id=\"fs-id1165042237957\">For this integral, let\u2019s choose [latex]u={\\tan}^{-1}x[\/latex] and [latex]dv=dx[\/latex], thereby making [latex]du=\\frac{1}{{x}^{2}+1}dx[\/latex] and [latex]v=x[\/latex]. After applying the integration-by-parts formula we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042238034\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan}^{-1}{x|}_{0}^{1}-\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042304278\">Use <em data-effect=\"italics\">u<\/em>-substitution to obtain<\/p>\r\n\r\n<div id=\"fs-id1165042304286\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx=\\frac{1}{2}\\text{ln}|{x}^{2}+1{|}_{0}^{1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042304368\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042304371\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan }^{-1}x{|}_{0}^{1}-\\frac{1}{2}\\mathrm{ln}|{x}^{2}+1|{|}_{0}^{1}=\\frac{\\pi }{4}-\\frac{1}{2}\\mathrm{ln}2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042304463\">At this point it might not be a bad idea to do a \"reality check\" on the reasonableness of our solution. Since [latex]\\frac{\\pi }{4}-\\frac{1}{2}\\text{ln}2\\approx 0.4388[\/latex], and from Figure 1 we expect our area to be slightly less than 0.5, this solution appears to be reasonable.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042058875\" data-type=\"problem\">\r\n<p id=\"fs-id1165042058881\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex], the <em data-effect=\"italics\">x<\/em>-axis, the <em data-effect=\"italics\">y<\/em>-axis, and the line [latex]x=1[\/latex] about the <em data-effect=\"italics\">y<\/em>-axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1165042058936\" data-type=\"solution\">\r\n<p id=\"fs-id1165042058938\">The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_01_002\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233752\/CNX_Calc_Figure_07_01_002.jpg\" alt=\"This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.\" width=\"731\" height=\"272\" data-media-type=\"image\/jpeg\" \/> Figure 2. We can use the shell method to find a volume of revolution.[\/caption]<\/figure>\r\n<p id=\"fs-id1165042058966\">To find the volume using shells, we must evaluate [latex]2\\pi {\\displaystyle\\int }_{0}^{1}x{e}^{\\text{-}x}dx[\/latex]. To do this, let [latex]u=x[\/latex] and [latex]dv={e}^{\\text{-}x}[\/latex]. These choices lead to [latex]du=dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}{e}^{\\text{-}x}=\\text{-}{e}^{\\text{-}x}[\/latex]. Substituting into the integration-by-parts for definite integrals formula, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042281368\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill \\text{Volume}&amp; =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}x{e}^{\\text{-}x}dx=2\\pi \\left(\\text{-}x{e}^{\\text{-}x}{|}_{0}^{1}+\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx\\right)\\hfill &amp; &amp; \\text{Use integration by parts}.\\hfill \\\\ &amp; =-2\\pi x{e}^{\\text{-}x}{|}_{0}^{1}-2\\pi {e}^{\\text{-}x}{|}_{0}^{1}\\hfill &amp; &amp; \\text{Evaluate}\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx=\\text{-}{e}^{\\text{-}x}{|}_{0}^{1}.\\hfill \\\\ &amp; =2\\pi -\\frac{4\\pi }{e}.\\hfill &amp; &amp; \\text{Evaluate and simplify}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042071427\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1165042071432\">Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius [latex]1[\/latex] and height of [latex]\\frac{1}{e}[\/latex] added to the volume of a cone of base radius [latex]1[\/latex] and height of [latex]1-\\frac{1}{3}[\/latex]. Consequently, the solid should have a volume a bit less than<\/p>\r\n\r\n<div id=\"fs-id1165042071470\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\pi {\\left(1\\right)}^{2}\\frac{1}{e}+\\left(\\frac{\\pi }{3}\\right){\\left(1\\right)}^{2}\\left(1-\\frac{1}{e}\\right)=\\frac{2\\pi }{3e}-\\frac{\\pi }{3}\\approx 1.8177[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042071569\">Since [latex]2\\pi -\\frac{4\\pi }{e}\\approx 1.6603[\/latex], we see that our calculated volume is reasonable.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311292[\/ohm_question]<\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Evaluate definite integrals using integration techniques including substitution, and integration by parts<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Substitution for Definite Integrals<\/h2>\n<p id=\"fs-id1170573387843\">Substitution can be used with definite integrals. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>substitution with definite integrals<\/h3>\n<p id=\"fs-id1170570999523\">Let [latex]u=g(x)[\/latex] and let [latex]{g}^{\\text{\u2032}}[\/latex] be continuous over an interval [latex]\\left[a,b\\right],[\/latex] and let [latex]f[\/latex] be continuous over the range of [latex]u=g(x).[\/latex] Then,<\/p>\n<div id=\"fs-id1170571189407\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(g(x)){g}^{\\prime }(x)dx={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587723\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587723\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573440252\">Let [latex]u=1+2{x}^{3},[\/latex] so [latex]du=6{x}^{2}dx.[\/latex]<\/p>\n<p>Since the original function includes one factor of [latex]x^2[\/latex]\u00a0and [latex]du=6{x}^{2}dx,[\/latex] multiply both sides of the <em>du<\/em> equation by [latex]\\frac{1}{6}.[\/latex]<\/p>\n<p>Then,<\/p>\n<div id=\"fs-id1170573362498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}du\\hfill & =\\hfill & 6{x}^{2}dx\\hfill \\\\ \\frac{1}{6}du\\hfill & =\\hfill & {x}^{2}dx.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571157085\">To adjust the limits of integration, note that when [latex]x=0,u=1+2(0)=1,[\/latex] and when [latex]x=1,u=1+2(1)=3.[\/latex]<\/p>\n<p>Then,<\/p>\n<div id=\"fs-id1170571303367\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du.[\/latex]<\/div>\n<p id=\"fs-id1170571285285\">Evaluating this expression, we get<\/p>\n<div id=\"fs-id1170571285289\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du\\hfill & =(\\frac{1}{6})(\\frac{{u}^{6}}{6}){|}_{1}^{3}\\hfill \\\\ & =\\frac{1}{36}\\left[{(3)}^{6}-{(1)}^{6}\\right]\\hfill \\\\ & =\\frac{182}{9}.\\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\/Ak_y3lsBNfE?controls=0&amp;start=519&amp;end=664&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\/5.5Substitution519to664_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.5 Substitution&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587724\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572587724\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571347206\">Let [latex]u=4{x}^{3}+3.[\/latex] Then, [latex]du=8xdx.[\/latex]<\/p>\n<p>To adjust the limits of integration, we note that when [latex]x=0,u=3,[\/latex] and when [latex]x=1,u=7.[\/latex]<\/p>\n<p>So our substitution gives,<\/p>\n<div id=\"fs-id1170571052992\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\\hfill & =\\frac{1}{8}{\\displaystyle\\int }_{3}^{7}{e}^{u}du\\hfill \\\\ \\\\ & =\\frac{1}{8}{e}^{u}{|}_{3}^{7}\\hfill \\\\ & =\\frac{{e}^{7}-{e}^{3}}{8}\\hfill \\\\ & \\approx 134.568.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm211747\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=211747&theme=lumen&iframe_resize_id=ohm211747&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<h2 data-type=\"title\">Integration by Parts for Definite Integrals<\/h2>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<div>\n<h3>integration by parts for definite integrals<\/h3>\n<p id=\"fs-id1165042231049\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives on [latex]\\left[a,b\\right][\/latex]. Then<\/p>\n<div style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}udv={uv|}_{a}^{b}-{\\displaystyle\\int }_{a}^{b}vdu[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042058875\" data-type=\"problem\">\u00a0Find the area of the region bounded above by the graph of [latex]y={\\tan}^{-1}x[\/latex] and below by the [latex]x[\/latex] -axis over the interval [latex]\\left[0,1\\right][\/latex].<\/div>\n<div id=\"fs-id1165042071427\" data-type=\"commentary\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Show Solution<\/button><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042237897\" data-type=\"solution\">\n<p id=\"fs-id1165042237899\">This region is shown in Figure 1. To find the area, we must evaluate [latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}{\\tan}^{-1}xdx[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_07_01_001\"><figcaption><\/figcaption><figure style=\"width: 562px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233750\/CNX_Calc_Figure_07_01_001.jpg\" alt=\"This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.\" width=\"562\" height=\"570\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. To find the area of the shaded region, we have to use integration by parts.<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1165042237957\">For this integral, let\u2019s choose [latex]u={\\tan}^{-1}x[\/latex] and [latex]dv=dx[\/latex], thereby making [latex]du=\\frac{1}{{x}^{2}+1}dx[\/latex] and [latex]v=x[\/latex]. After applying the integration-by-parts formula we obtain<\/p>\n<div id=\"fs-id1165042238034\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan}^{-1}{x|}_{0}^{1}-\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042304278\">Use <em data-effect=\"italics\">u<\/em>-substitution to obtain<\/p>\n<div id=\"fs-id1165042304286\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx=\\frac{1}{2}\\text{ln}|{x}^{2}+1{|}_{0}^{1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042304368\">Thus,<\/p>\n<div id=\"fs-id1165042304371\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{Area}=x{\\tan }^{-1}x{|}_{0}^{1}-\\frac{1}{2}\\mathrm{ln}|{x}^{2}+1|{|}_{0}^{1}=\\frac{\\pi }{4}-\\frac{1}{2}\\mathrm{ln}2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042304463\">At this point it might not be a bad idea to do a &#8220;reality check&#8221; on the reasonableness of our solution. Since [latex]\\frac{\\pi }{4}-\\frac{1}{2}\\text{ln}2\\approx 0.4388[\/latex], and from Figure 1 we expect our area to be slightly less than 0.5, this solution appears to be reasonable.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div data-type=\"problem\">\n<p id=\"fs-id1165042058881\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex], the <em data-effect=\"italics\">x<\/em>-axis, the <em data-effect=\"italics\">y<\/em>-axis, and the line [latex]x=1[\/latex] about the <em data-effect=\"italics\">y<\/em>-axis.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558869\">Show Solution<\/button><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042058936\" data-type=\"solution\">\n<p id=\"fs-id1165042058938\">The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).<\/p>\n<figure id=\"CNX_Calc_Figure_07_01_002\"><figcaption><\/figcaption><figure style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233752\/CNX_Calc_Figure_07_01_002.jpg\" alt=\"This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.\" width=\"731\" height=\"272\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 2. We can use the shell method to find a volume of revolution.<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1165042058966\">To find the volume using shells, we must evaluate [latex]2\\pi {\\displaystyle\\int }_{0}^{1}x{e}^{\\text{-}x}dx[\/latex]. To do this, let [latex]u=x[\/latex] and [latex]dv={e}^{\\text{-}x}[\/latex]. These choices lead to [latex]du=dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}{e}^{\\text{-}x}=\\text{-}{e}^{\\text{-}x}[\/latex]. Substituting into the integration-by-parts for definite integrals formula, we obtain<\/p>\n<div id=\"fs-id1165042281368\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill \\text{Volume}& =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}x{e}^{\\text{-}x}dx=2\\pi \\left(\\text{-}x{e}^{\\text{-}x}{|}_{0}^{1}+\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx\\right)\\hfill & & \\text{Use integration by parts}.\\hfill \\\\ & =-2\\pi x{e}^{\\text{-}x}{|}_{0}^{1}-2\\pi {e}^{\\text{-}x}{|}_{0}^{1}\\hfill & & \\text{Evaluate}\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx=\\text{-}{e}^{\\text{-}x}{|}_{0}^{1}.\\hfill \\\\ & =2\\pi -\\frac{4\\pi }{e}.\\hfill & & \\text{Evaluate and simplify}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1165042071432\">Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius [latex]1[\/latex] and height of [latex]\\frac{1}{e}[\/latex] added to the volume of a cone of base radius [latex]1[\/latex] and height of [latex]1-\\frac{1}{3}[\/latex]. Consequently, the solid should have a volume a bit less than<\/p>\n<div id=\"fs-id1165042071470\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\pi {\\left(1\\right)}^{2}\\frac{1}{e}+\\left(\\frac{\\pi }{3}\\right){\\left(1\\right)}^{2}\\left(1-\\frac{1}{e}\\right)=\\frac{2\\pi }{3e}-\\frac{\\pi }{3}\\approx 1.8177[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042071569\">Since [latex]2\\pi -\\frac{4\\pi }{e}\\approx 1.6603[\/latex], we see that our calculated volume is reasonable.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311292\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311292&theme=lumen&iframe_resize_id=ohm311292&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":2,"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-673","chapter","type-chapter","status-publish","hentry"],"part":432,"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\/673","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\/673\/revisions"}],"predecessor-version":[{"id":2657,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/673\/revisions\/2657"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/432"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/673\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=673"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=673"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=673"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=673"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}