{"id":1612,"date":"2021-07-22T16:56:39","date_gmt":"2021-07-22T16:56:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1612"},"modified":"2022-03-19T04:22:04","modified_gmt":"2022-03-19T04:22:04","slug":"evaluating-improper-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/evaluating-improper-integrals\/","title":{"raw":"Evaluating Improper Integrals","rendered":"Evaluating Improper Integrals"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate an integral over an infinite interval<\/li>\r\n \t<li>Evaluate an integral over a closed interval with an infinite discontinuity within the interval<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1165043219173\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Integrating over an Infinite Interval<\/h2>\r\n<p id=\"fs-id1165043161272\">How should we go about defining an integral of the type [latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx?[\/latex] We can integrate [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] for any value of [latex]t[\/latex], so it is reasonable to look at the behavior of this integral as we substitute larger values of [latex]t[\/latex]. Figure 1 shows that [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] may be interpreted as area for various values of [latex]t[\/latex]. In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_07_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233844\/CNX_Calc_Figure_07_07_001.jpg\" alt=\"This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and decreasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. The region in the first curve is small, and progressively gets wider under the second and third graph as t moves further to the right away from a on the x-axis.\" width=\"975\" height=\"165\" data-media-type=\"image\/jpeg\" \/> Figure 1. To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.[\/caption]<\/figure>\r\n<div id=\"fs-id1165043423093\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<ol id=\"fs-id1165042516485\" type=\"1\">\r\n \t<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over an interval of the form [latex]\\left[a,\\text{+}\\infty \\right)[\/latex]. Then<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165043372378\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nprovided this limit exists.<\/li>\r\n \t<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over an interval of the form [latex]\\left(\\text{-}\\infty ,b\\right][\/latex]. Then<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165043086270\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{b}f\\left(x\\right)dx=\\underset{t\\to \\text{-}\\infty }{\\text{lim}}{\\displaystyle\\int }_{t}^{b}f\\left(x\\right)dx[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nprovided this limit exists.<span data-type=\"newline\">\r\n<\/span>\r\nIn each case, if the limit exists, then the <strong>improper integral<\/strong> is said to converge. If the limit does not exist, then the improper integral is said to diverge.<\/li>\r\n \t<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over [latex]\\left(\\text{-}\\infty ,\\text{+}\\infty \\right)[\/latex]. Then<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042478832\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }f\\left(x\\right)dx={\\displaystyle\\int }_{\\text{-}\\infty }^{0}f\\left(x\\right)dx+{\\displaystyle\\int }_{0}^{+\\infty }f\\left(x\\right)dx[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nprovided that [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}f\\left(x\\right)dx[\/latex] and [latex]{\\displaystyle\\int }_{0}^{+\\infty }f\\left(x\\right)dx[\/latex] both converge. If either of these two integrals diverge, then [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }f\\left(x\\right)dx[\/latex] diverges. (It can be shown that, in fact, [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }f\\left(x\\right)dx={\\displaystyle\\int }_{\\text{-}\\infty }^{a}f\\left(x\\right)dx+{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx[\/latex] for any value of [latex]a.[\/latex])<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165043014225\">In our first example, we return to the question we posed at the start of this section: Is the area between the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the [latex]x[\/latex] -axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] finite or infinite?<\/p>\r\n\r\n<div id=\"fs-id1165043428127\" data-type=\"example\">\r\n<div id=\"fs-id1165043001948\" data-type=\"exercise\">\r\n<div id=\"fs-id1165043085741\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding an Area<\/h3>\r\n<div id=\"fs-id1165043085741\" data-type=\"problem\">\r\n<p id=\"fs-id1165043346878\">Determine whether the area between the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is finite or infinite.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1165042330814\" data-type=\"solution\">\r\n<p id=\"fs-id1165043350067\">We first do a quick sketch of the region in question, as shown in the following graph.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_07_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233847\/CNX_Calc_Figure_07_07_002.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. In the first quadrant there is a shaded region under the curve bounded by x = 1 and x = 4.\" width=\"487\" height=\"351\" data-media-type=\"image\/jpeg\" \/> Figure 2. We can find the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the x-axis on an infinite interval.[\/caption]<\/figure>\r\n<p id=\"fs-id1165042925668\">We can see that the area of this region is given by [latex]A={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx[\/latex]. Then we have<\/p>\r\n\r\n<div id=\"fs-id1165043131589\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill A&amp; ={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx\\hfill &amp; &amp; &amp; \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{x}dx\\hfill &amp; &amp; &amp; \\text{Rewrite the improper integral as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\text{ln}|x||{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{ln}|t|-\\text{ln}1\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\text{+}\\infty .\\hfill &amp; &amp; &amp; \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043117138\">Since the improper integral diverges to [latex]+\\infty [\/latex], the area of the region is infinite.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042988868\" data-type=\"example\">\r\n<div id=\"fs-id1165042514822\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042925771\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding a Volume<\/h3>\r\n<div id=\"fs-id1165042925771\" data-type=\"problem\">\r\n<p id=\"fs-id1165042677398\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] about the [latex]x[\/latex] -axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1165042582834\" data-type=\"solution\">\r\n<p id=\"fs-id1165042978442\">The solid is shown in Figure 3. Using the disk method, we see that the volume <em data-effect=\"italics\">V<\/em> is<\/p>\r\n\r\n<div id=\"fs-id1165042321060\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]V=\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<figure id=\"CNX_Calc_Figure_07_07_003\"><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\/11233850\/CNX_Calc_Figure_07_07_003.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. The graph shows a solid that has been generated by rotating the curve in the first quadrant around the x-axis.\" width=\"731\" height=\"351\" data-media-type=\"image\/jpeg\" \/> Figure 3. The solid of revolution can be generated by rotating an infinite area about the x-axis.[\/caption]<\/figure>\r\n<p id=\"fs-id1165043112190\">Then we have<\/p>\r\n\r\n<div id=\"fs-id1165043259904\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill V&amp; =\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx\\hfill &amp; &amp; &amp; \\\\ &amp; =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{{x}^{2}}dx\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}-\\frac{1}{x}|{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}\\frac{1}{t}+1\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\pi .\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042788783\">The improper integral converges to [latex]\\pi [\/latex]. Therefore, the volume of the solid of revolution is [latex]\\pi [\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165043348520\">In conclusion, although the area of the region between the <em data-effect=\"italics\">x<\/em>-axis and the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is infinite, the volume of the solid generated by revolving this region about the <em data-effect=\"italics\">x<\/em>-axis is finite. The solid generated is known as <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">Gabriel\u2019s Horn<\/em><\/span>.<\/p>\r\n\r\n<div id=\"fs-id1165043122367\" class=\"media-2\" data-type=\"note\">\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1165043182674\">Visit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Gabriel%27s_Horn\" target=\"_blank\" rel=\"noopener\">this website to read more about Gabriel\u2019s Horn<\/a>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042804162\" data-type=\"example\">\r\n<div id=\"fs-id1165043109202\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042988255\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Evaluating an Improper Integral over an Infinite Interval<\/h3>\r\n<div id=\"fs-id1165042988255\" data-type=\"problem\">\r\n<p id=\"fs-id1165043373560\">Evaluate [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}\\frac{1}{{x}^{2}+4}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1165043097500\" data-type=\"solution\">\r\n<p id=\"fs-id1165042534998\">Begin by rewriting [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}\\frac{1}{{x}^{2}+4}dx[\/latex] as a limit using the equation 2 from the definition. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042316105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{\\text{-}\\infty }^{0}\\frac{1}{{x}^{2}+4}dx&amp; =\\underset{x\\to \\text{-}\\infty }{\\text{lim}}{\\displaystyle\\int }_{t}^{0}\\frac{1}{{x}^{2}+4}dx\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{-}\\infty }{\\text{lim}}\\frac{1}{2}{\\tan}^{-1}\\frac{x}{2}|{}_{\\begin{array}{c}\\\\ t\\end{array}}^{\\begin{array}{c}0\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\frac{1}{2}\\underset{t\\to \\text{-}\\infty }{\\text{lim}}\\left({\\tan}^{-1}0-{\\tan}^{-1}\\frac{t}{2}\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\frac{\\pi }{4}.\\hfill &amp; &amp; &amp; \\text{Evaluate the limit and simplify.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043422657\">The improper integral converges to [latex]\\frac{\\pi }{4}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nBecause improper integrals require evaluating limits at infinity, at times we may be required to use L'H\u00f4pital's Rule to evaluate a limit.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: L'H\u00f4pital's Rule<\/h3>\r\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval [latex]\\left(a, \\infty \\right) [\/latex] for some value of [latex] a [\/latex]. If either:<\/p>\r\n\r\n<ol>\r\n \t<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to \\infty}{\\lim}g(x)=0[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)= \\infty [\/latex] (or [latex] -\\infty [\/latex]) and\u00a0[latex]\\underset{x\\to \\infty}{\\lim}g(x)= \\infty [\/latex] (or [latex] -\\infty [\/latex]), then<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to \\infty}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042508139\" data-type=\"example\">\r\n<div id=\"fs-id1165042608025\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042608027\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Evaluating an Improper Integral on [latex]\\left(-\\infty ,+\\infty \\right)[\/latex]<\/h3>\r\n<div id=\"fs-id1165042608027\" data-type=\"problem\">\r\n<p id=\"fs-id1165042318435\">Evaluate [latex]{\\displaystyle\\int }_{-\\infty }^{+\\infty }x{e}^{x}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1165042960104\" data-type=\"solution\">\r\n<p id=\"fs-id1165042705512\">Start by splitting up the integral:<\/p>\r\n\r\n<div id=\"fs-id1165042705515\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx={\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx+{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042519801\">If either [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx[\/latex] or [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges, then [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges. Compute each integral separately. For the first integral,<\/p>\r\n\r\n<div id=\"fs-id1165039560750\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{-\\infty }^{0}x{e}^{x}dx}&amp; ={\\underset{t\\to -\\infty}\\lim}{\\displaystyle\\int _{t}^{0}x{e}^{x}dx}\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; ={\\underset{t\\to -\\infty}\\lim}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Use integration by parts to find the} \\hfill \\\\ \\text{antiderivative. (Here } u=x \\text{ and } dv=e_{dv}^{x}\\text{.)}\\end{array} \\\\ &amp; ={\\underset{t\\to -\\infty}\\lim}\\left(-1-t{e}^{t}+{e}^{t}\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =-1.\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Evaluate the limit.}\\mathit{\\text{Note:}} {\\underset{t\\to -\\infty}\\lim}t{e}^{t}\\text{is}\\hfill \\\\ \\text{indeterminate of the form}0\\cdot \\infty .\\text{Thus,}\\hfill \\\\ {\\underset{t\\to -\\infty}\\lim}t{e}^{t}={\\underset{t\\to -\\infty}\\lim}\\frac{t}{{e}^{\\text{-}t}}={\\underset{t\\to -\\infty}\\lim}\\frac{-1}{{e}^{-t}}={\\underset{t\\to -\\infty}\\lim}-{e}^{t}=0\\text{by}\\hfill \\\\ \\text{L'H}\\hat{o}\\text{pital's Rule.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043354391\">The first improper integral converges. For the second integral,<\/p>\r\n\r\n<div id=\"fs-id1165043354394\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{0}^{+\\infty }x{e}^{x}dx}&amp; =\\underset{t\\to +\\infty }{\\text{lim}}{\\displaystyle\\int _{0}^{t}x{e}^{x}dx}\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(t{e}^{t}-{e}^{t}+1\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\left(t - 1\\right){e}^{t}+1\\right)\\hfill &amp; &amp; &amp; \\text{Rewrite.} (t e^t-e^t \\text{ is indeterminate.)} \\hfill \\\\ &amp; =\\text{+}\\infty .\\hfill &amp; &amp; &amp; \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042910269\">Thus, [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges. Since this integral diverges, [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges as well.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043096729\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165043096733\" data-type=\"exercise\">\r\n<div id=\"fs-id1165043103020\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165043103020\" data-type=\"problem\">\r\n<p id=\"fs-id1165043103022\">Evaluate [latex]{\\displaystyle\\int }_{-3}^{+\\infty }{e}^{\\text{-}x}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1165043272721\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165042706105\">[latex]{\\displaystyle\\int }_{-3}^{+\\infty }{e}^{\\text{-}x}dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{-3}^{t}{e}^{\\text{-}x}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1165042355763\" data-type=\"solution\">\r\n<p id=\"fs-id1165042355765\">[latex]{e}^{3}[\/latex], converges<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8QzQO748O58?controls=0&amp;start=676&amp;end=750&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">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>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.7ImproperIntegrals676to750_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.7 Improper Integrals\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section id=\"fs-id1165042543169\" data-depth=\"1\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]25530[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Integrating a Discontinuous Integrand<\/h2>\r\n<p id=\"fs-id1165042543174\">Now let\u2019s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex], where [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right)[\/latex] and discontinuous at [latex]b[\/latex]. Since the function [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,t\\right][\/latex] for all values of [latex]t[\/latex] satisfying [latex]a&lt;t&lt;b[\/latex], the integral [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] is defined for all such values of [latex]t[\/latex]. Thus, it makes sense to consider the values of [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] as [latex]t[\/latex] approaches [latex]b[\/latex] for [latex]a&lt;t&lt;b[\/latex]. That is, we define [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx=\\underset{t\\to {b}^{-}}{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex], provided this limit exists. Figure 4 illustrates [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] as areas of regions for values of [latex]t[\/latex] approaching [latex]b[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_07_004\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233856\/CNX_Calc_Figure_07_07_004.jpg\" alt=\"This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and increasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. There is also a vertical asymptote at x = b. The region in the first curve is small, and progressively gets wider under the second and third graph as t gets further from a, and closer to b on the x-axis.\" width=\"975\" height=\"239\" data-media-type=\"image\/jpeg\" \/> Figure 4. As [latex]t[\/latex] approaches b from the left, the value of the area from a to [latex]t[\/latex] approaches the area from a to b.[\/caption]<\/figure>\r\n<p id=\"fs-id1165043094066\">We use a similar approach to define [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex], where [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left(a,b\\right][\/latex] and discontinuous at [latex]a[\/latex]. We now proceed with a formal definition.<\/p>\r\n\r\n<div id=\"fs-id1165042333401\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<ol id=\"fs-id1165042333405\" type=\"1\">\r\n \t<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over [latex]\\left[a,b\\right)[\/latex]. Then,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042617693\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx=\\underset{t\\to {b}^{-}}{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over [latex]\\left(a,b\\right][\/latex]. Then,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042506827\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx=\\underset{t\\to {a}^{+}}{\\text{lim}}{\\displaystyle\\int }_{t}^{b}f\\left(x\\right)dx[\/latex].<\/div>\r\nIn each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.<\/li>\r\n \t<li>If [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right][\/latex] except at a point [latex]c[\/latex] in [latex]\\left(a,b\\right)[\/latex], then<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165043378015\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx={\\displaystyle\\int }_{a}^{c}f\\left(x\\right)dx+{\\displaystyle\\int }_{c}^{b}f\\left(x\\right)dx[\/latex],<\/div>\r\nprovided both [latex]{\\displaystyle\\int }_{a}^{c}f\\left(x\\right)dx[\/latex] and [latex]{\\displaystyle\\int }_{c}^{b}f\\left(x\\right)dx[\/latex] converge. If either of these integrals diverges, then [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex] diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165043257647\">The following examples demonstrate the application of this definition.<\/p>\r\n\r\n<div id=\"fs-id1165043257650\" data-type=\"example\">\r\n<div id=\"fs-id1165043257652\" data-type=\"exercise\">\r\n<div id=\"fs-id1165043257655\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating a Discontinuous Integrand<\/h3>\r\n<div id=\"fs-id1165043257655\" data-type=\"problem\">\r\n<p id=\"fs-id1165043119755\">Evaluate [latex]{\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx[\/latex], if possible. State whether the integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1165042364402\" data-type=\"solution\">\r\n<p id=\"fs-id1165042364404\">The function [latex]f\\left(x\\right)=\\frac{1}{\\sqrt{4-x}}[\/latex] is continuous over [latex]\\left[0,4\\right)[\/latex] and discontinuous at 4. Using equation 1 from the definition, rewrite [latex]{\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx[\/latex] as a limit:<\/p>\r\n\r\n<div id=\"fs-id1165043373976\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx&amp; =\\underset{t\\to {4}^{-}}{\\text{lim}}{\\displaystyle\\int }_{0}^{t}\\frac{1}{\\sqrt{4-x}}dx\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to {4}^{-}}{\\text{lim}}\\left(-2\\sqrt{4-x}\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to {4}^{-}}{\\text{lim}}\\left(-2\\sqrt{4-t}+4\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =4.\\hfill &amp; &amp; &amp; \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043173672\">The improper integral converges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043173678\" data-type=\"example\">\r\n<div id=\"fs-id1165043173681\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042375631\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating a Discontinuous Integrand<\/h3>\r\n<div id=\"fs-id1165042375631\" data-type=\"problem\">\r\n<p id=\"fs-id1165042375636\">Evaluate [latex]{\\displaystyle\\int }_{0}^{2}x\\text{ln}xdx[\/latex]. State whether the integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1165042538864\" data-type=\"solution\">\r\n<p id=\"fs-id1165042538867\">Since [latex]f\\left(x\\right)=x\\ln{x}[\/latex] is continuous over [latex]\\left(0,2\\right][\/latex] and is discontinuous at zero, we can rewrite the integral in limit form using equation 2 from the definition:<\/p>\r\n\r\n<div id=\"fs-id1165042865839\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{2}x\\text{ln}xdx&amp; =\\underset{t\\to {0}^{+}}{\\text{lim}}{\\displaystyle\\int }_{t}^{2}x\\text{ln}xdx\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to {0}^{+}}{\\text{lim}}\\left(\\frac{1}{2}{x}^{2}\\text{ln}x-\\frac{1}{4}{x}^{2}\\right)|{}_{\\begin{array}{c}\\\\ t\\end{array}}^{\\begin{array}{c}2\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Evaluate}{\\displaystyle\\int}x\\ln{x}dx\\text{ using integration by parts}\\hfill \\\\ \\text{with }u=\\text{ln}x\\text{ and }dv=x.\\hfill \\end{array}\\hfill \\\\ &amp; =\\underset{t\\to {0}^{+}}{\\text{lim}}\\left(2\\text{ln}2 - 1-\\frac{1}{2}{t}^{2}\\text{ln}t+\\frac{1}{4}{t}^{2}\\right).\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =2\\text{ln}2 - 1.\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Evaluate the limit.}\\underset{t\\to {0}^{+}}{\\text{lim}}{t}^{2}\\ln{t}\\text{ is indeterminate.}\\hfill \\\\ \\text{To evaluate it, rewrite as a quotient and apply}\\hfill \\\\ \\text{L'h}\\hat{o}\\text{pital's rule.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043423544\">The improper integral converges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043423550\" data-type=\"example\">\r\n<div id=\"fs-id1165043423552\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042396199\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating a Discontinuous Integrand<\/h3>\r\n<div id=\"fs-id1165042396199\" data-type=\"problem\">\r\n<p id=\"fs-id1165042396204\">Evaluate [latex]{\\displaystyle\\int }_{-1}^{1}\\frac{1}{{x}^{3}}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1165042321413\" data-type=\"solution\">\r\n<p id=\"fs-id1165042321415\">Since [latex]f\\left(x\\right)=\\frac{1}{{x}^{3}}[\/latex] is discontinuous at zero, using equation 3 from the definition, we can write<\/p>\r\n\r\n<div id=\"fs-id1165042713598\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{-1}^{1}\\frac{1}{{x}^{3}}dx={\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx+{\\displaystyle\\int }_{0}^{1}\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042985556\">If either of the two integrals diverges, then the original integral diverges. Begin with [latex]{\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx:[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165042645708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx&amp; =\\underset{t\\to {0}^{-}}{\\text{lim}}{\\displaystyle\\int }_{-1}^{t}\\frac{1}{{x}^{3}}dx\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to {0}^{-}}{\\text{lim}}\\left(-\\frac{1}{2{x}^{2}}\\right)|{}_{\\begin{array}{c}\\\\ -1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to {0}^{-}}{\\text{lim}}\\left(-\\frac{1}{2{t}^{2}}+\\frac{1}{2}\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\text{+}\\infty .\\hfill &amp; &amp; &amp; \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043430752\">Therefore, [latex]{\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx[\/latex] diverges. Since [latex]{\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx[\/latex] diverges, [latex]{\\displaystyle\\int }_{-1}^{1}\\frac{1}{{x}^{3}}dx[\/latex] diverges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042376188\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042376192\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042376194\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165042376194\" data-type=\"problem\">\r\n<p id=\"fs-id1165042376196\">Evaluate [latex]{\\displaystyle\\int }_{0}^{2}\\frac{1}{x}dx[\/latex]. State whether the integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1165042422655\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165042708426\">Write [latex]{\\displaystyle\\int }_{0}^{2}\\frac{1}{x}dx[\/latex] in limit form using equation 2 from the definition.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1165042422640\" data-type=\"solution\">\r\n<p id=\"fs-id1165042422642\">[latex]+\\infty [\/latex], diverges<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8QzQO748O58?controls=0&amp;start=1388&amp;end=1468&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">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>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.7ImproperIntegrals1388to1468_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.7 Improper Integrals\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section id=\"fs-id1165043273047\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate an integral over an infinite interval<\/li>\n<li>Evaluate an integral over a closed interval with an infinite discontinuity within the interval<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1165043219173\" data-depth=\"1\">\n<h2 data-type=\"title\">Integrating over an Infinite Interval<\/h2>\n<p id=\"fs-id1165043161272\">How should we go about defining an integral of the type [latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx?[\/latex] We can integrate [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] for any value of [latex]t[\/latex], so it is reasonable to look at the behavior of this integral as we substitute larger values of [latex]t[\/latex]. Figure 1 shows that [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] may be interpreted as area for various values of [latex]t[\/latex]. In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_001\"><figcaption><\/figcaption><div style=\"width: 985px\" 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\/11233844\/CNX_Calc_Figure_07_07_001.jpg\" alt=\"This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and decreasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. The region in the first curve is small, and progressively gets wider under the second and third graph as t moves further to the right away from a on the x-axis.\" width=\"975\" height=\"165\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1165043423093\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<ol id=\"fs-id1165042516485\" type=\"1\">\n<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over an interval of the form [latex]\\left[a,\\text{+}\\infty \\right)[\/latex]. Then<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165043372378\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nprovided this limit exists.<\/li>\n<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over an interval of the form [latex]\\left(\\text{-}\\infty ,b\\right][\/latex]. Then<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165043086270\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{b}f\\left(x\\right)dx=\\underset{t\\to \\text{-}\\infty }{\\text{lim}}{\\displaystyle\\int }_{t}^{b}f\\left(x\\right)dx[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nprovided this limit exists.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIn each case, if the limit exists, then the <strong>improper integral<\/strong> is said to converge. If the limit does not exist, then the improper integral is said to diverge.<\/li>\n<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over [latex]\\left(\\text{-}\\infty ,\\text{+}\\infty \\right)[\/latex]. Then<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042478832\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }f\\left(x\\right)dx={\\displaystyle\\int }_{\\text{-}\\infty }^{0}f\\left(x\\right)dx+{\\displaystyle\\int }_{0}^{+\\infty }f\\left(x\\right)dx[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nprovided that [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}f\\left(x\\right)dx[\/latex] and [latex]{\\displaystyle\\int }_{0}^{+\\infty }f\\left(x\\right)dx[\/latex] both converge. If either of these two integrals diverge, then [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }f\\left(x\\right)dx[\/latex] diverges. (It can be shown that, in fact, [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }f\\left(x\\right)dx={\\displaystyle\\int }_{\\text{-}\\infty }^{a}f\\left(x\\right)dx+{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx[\/latex] for any value of [latex]a.[\/latex])<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165043014225\">In our first example, we return to the question we posed at the start of this section: Is the area between the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the [latex]x[\/latex] -axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] finite or infinite?<\/p>\n<div id=\"fs-id1165043428127\" data-type=\"example\">\n<div id=\"fs-id1165043001948\" data-type=\"exercise\">\n<div id=\"fs-id1165043085741\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding an Area<\/h3>\n<div id=\"fs-id1165043085741\" data-type=\"problem\">\n<p id=\"fs-id1165043346878\">Determine whether the area between the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is finite or infinite.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042330814\" data-type=\"solution\">\n<p id=\"fs-id1165043350067\">We first do a quick sketch of the region in question, as shown in the following graph.<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_002\"><figcaption><\/figcaption><div style=\"width: 497px\" 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\/11233847\/CNX_Calc_Figure_07_07_002.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. In the first quadrant there is a shaded region under the curve bounded by x = 1 and x = 4.\" width=\"487\" height=\"351\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. We can find the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the x-axis on an infinite interval.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165042925668\">We can see that the area of this region is given by [latex]A={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx[\/latex]. Then we have<\/p>\n<div id=\"fs-id1165043131589\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill A& ={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx\\hfill & & & \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{x}dx\\hfill & & & \\text{Rewrite the improper integral as a limit.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\text{ln}|x||{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{ln}|t|-\\text{ln}1\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\text{+}\\infty .\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043117138\">Since the improper integral diverges to [latex]+\\infty[\/latex], the area of the region is infinite.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042988868\" data-type=\"example\">\n<div id=\"fs-id1165042514822\" data-type=\"exercise\">\n<div id=\"fs-id1165042925771\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding a Volume<\/h3>\n<div id=\"fs-id1165042925771\" data-type=\"problem\">\n<p id=\"fs-id1165042677398\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] about the [latex]x[\/latex] -axis.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042582834\" data-type=\"solution\">\n<p id=\"fs-id1165042978442\">The solid is shown in Figure 3. Using the disk method, we see that the volume <em data-effect=\"italics\">V<\/em> is<\/p>\n<div id=\"fs-id1165042321060\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]V=\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_003\"><figcaption><\/figcaption><div style=\"width: 741px\" 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\/11233850\/CNX_Calc_Figure_07_07_003.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. The graph shows a solid that has been generated by rotating the curve in the first quadrant around the x-axis.\" width=\"731\" height=\"351\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The solid of revolution can be generated by rotating an infinite area about the x-axis.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165043112190\">Then we have<\/p>\n<div id=\"fs-id1165043259904\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill V& =\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx\\hfill & & & \\\\ & =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{{x}^{2}}dx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}-\\frac{1}{x}|{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}\\frac{1}{t}+1\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\pi .\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042788783\">The improper integral converges to [latex]\\pi[\/latex]. Therefore, the volume of the solid of revolution is [latex]\\pi[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165043348520\">In conclusion, although the area of the region between the <em data-effect=\"italics\">x<\/em>-axis and the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is infinite, the volume of the solid generated by revolving this region about the <em data-effect=\"italics\">x<\/em>-axis is finite. The solid generated is known as <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">Gabriel\u2019s Horn<\/em><\/span>.<\/p>\n<div id=\"fs-id1165043122367\" class=\"media-2\" data-type=\"note\">\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1165043182674\">Visit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Gabriel%27s_Horn\" target=\"_blank\" rel=\"noopener\">this website to read more about Gabriel\u2019s Horn<\/a>.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042804162\" data-type=\"example\">\n<div id=\"fs-id1165043109202\" data-type=\"exercise\">\n<div id=\"fs-id1165042988255\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Evaluating an Improper Integral over an Infinite Interval<\/h3>\n<div id=\"fs-id1165042988255\" data-type=\"problem\">\n<p id=\"fs-id1165043373560\">Evaluate [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}\\frac{1}{{x}^{2}+4}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Show Solution<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165043097500\" data-type=\"solution\">\n<p id=\"fs-id1165042534998\">Begin by rewriting [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}\\frac{1}{{x}^{2}+4}dx[\/latex] as a limit using the equation 2 from the definition. Thus,<\/p>\n<div id=\"fs-id1165042316105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{\\text{-}\\infty }^{0}\\frac{1}{{x}^{2}+4}dx& =\\underset{x\\to \\text{-}\\infty }{\\text{lim}}{\\displaystyle\\int }_{t}^{0}\\frac{1}{{x}^{2}+4}dx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to \\text{-}\\infty }{\\text{lim}}\\frac{1}{2}{\\tan}^{-1}\\frac{x}{2}|{}_{\\begin{array}{c}\\\\ t\\end{array}}^{\\begin{array}{c}0\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\frac{1}{2}\\underset{t\\to \\text{-}\\infty }{\\text{lim}}\\left({\\tan}^{-1}0-{\\tan}^{-1}\\frac{t}{2}\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\frac{\\pi }{4}.\\hfill & & & \\text{Evaluate the limit and simplify.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043422657\">The improper integral converges to [latex]\\frac{\\pi }{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Because improper integrals require evaluating limits at infinity, at times we may be required to use L&#8217;H\u00f4pital&#8217;s Rule to evaluate a limit.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: L&#8217;H\u00f4pital&#8217;s Rule<\/h3>\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval [latex]\\left(a, \\infty \\right)[\/latex] for some value of [latex]a[\/latex]. If either:<\/p>\n<ol>\n<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to \\infty}{\\lim}g(x)=0[\/latex]<\/li>\n<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)= \\infty[\/latex] (or [latex]-\\infty[\/latex]) and\u00a0[latex]\\underset{x\\to \\infty}{\\lim}g(x)= \\infty[\/latex] (or [latex]-\\infty[\/latex]), then<\/li>\n<\/ol>\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to \\infty}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165042508139\" data-type=\"example\">\n<div id=\"fs-id1165042608025\" data-type=\"exercise\">\n<div id=\"fs-id1165042608027\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Evaluating an Improper Integral on [latex]\\left(-\\infty ,+\\infty \\right)[\/latex]<\/h3>\n<div id=\"fs-id1165042608027\" data-type=\"problem\">\n<p id=\"fs-id1165042318435\">Evaluate [latex]{\\displaystyle\\int }_{-\\infty }^{+\\infty }x{e}^{x}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042960104\" data-type=\"solution\">\n<p id=\"fs-id1165042705512\">Start by splitting up the integral:<\/p>\n<div id=\"fs-id1165042705515\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx={\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx+{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042519801\">If either [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx[\/latex] or [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges, then [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges. Compute each integral separately. For the first integral,<\/p>\n<div id=\"fs-id1165039560750\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{-\\infty }^{0}x{e}^{x}dx}& ={\\underset{t\\to -\\infty}\\lim}{\\displaystyle\\int _{t}^{0}x{e}^{x}dx}\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & ={\\underset{t\\to -\\infty}\\lim}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill & & & \\begin{array}{c}\\text{Use integration by parts to find the} \\hfill \\\\ \\text{antiderivative. (Here } u=x \\text{ and } dv=e_{dv}^{x}\\text{.)}\\end{array} \\\\ & ={\\underset{t\\to -\\infty}\\lim}\\left(-1-t{e}^{t}+{e}^{t}\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =-1.\\hfill & & & \\begin{array}{c}\\text{Evaluate the limit.}\\mathit{\\text{Note:}} {\\underset{t\\to -\\infty}\\lim}t{e}^{t}\\text{is}\\hfill \\\\ \\text{indeterminate of the form}0\\cdot \\infty .\\text{Thus,}\\hfill \\\\ {\\underset{t\\to -\\infty}\\lim}t{e}^{t}={\\underset{t\\to -\\infty}\\lim}\\frac{t}{{e}^{\\text{-}t}}={\\underset{t\\to -\\infty}\\lim}\\frac{-1}{{e}^{-t}}={\\underset{t\\to -\\infty}\\lim}-{e}^{t}=0\\text{by}\\hfill \\\\ \\text{L'H}\\hat{o}\\text{pital's Rule.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043354391\">The first improper integral converges. For the second integral,<\/p>\n<div id=\"fs-id1165043354394\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{0}^{+\\infty }x{e}^{x}dx}& =\\underset{t\\to +\\infty }{\\text{lim}}{\\displaystyle\\int _{0}^{t}x{e}^{x}dx}\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(t{e}^{t}-{e}^{t}+1\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\left(t - 1\\right){e}^{t}+1\\right)\\hfill & & & \\text{Rewrite.} (t e^t-e^t \\text{ is indeterminate.)} \\hfill \\\\ & =\\text{+}\\infty .\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042910269\">Thus, [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges. Since this integral diverges, [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges as well.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043096729\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165043096733\" data-type=\"exercise\">\n<div id=\"fs-id1165043103020\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165043103020\" data-type=\"problem\">\n<p id=\"fs-id1165043103022\">Evaluate [latex]{\\displaystyle\\int }_{-3}^{+\\infty }{e}^{\\text{-}x}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165043272721\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165042706105\">[latex]{\\displaystyle\\int }_{-3}^{+\\infty }{e}^{\\text{-}x}dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{-3}^{t}{e}^{\\text{-}x}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042355763\" data-type=\"solution\">\n<p id=\"fs-id1165042355765\">[latex]{e}^{3}[\/latex], converges<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8QzQO748O58?controls=0&amp;start=676&amp;end=750&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">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+II\/Transcripts\/3.7ImproperIntegrals676to750_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.7 Improper Integrals&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section id=\"fs-id1165042543169\" data-depth=\"1\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm25530\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=25530&theme=oea&iframe_resize_id=ohm25530&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 data-type=\"title\">Integrating a Discontinuous Integrand<\/h2>\n<p id=\"fs-id1165042543174\">Now let\u2019s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex], where [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right)[\/latex] and discontinuous at [latex]b[\/latex]. Since the function [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,t\\right][\/latex] for all values of [latex]t[\/latex] satisfying [latex]a<t<b[\/latex], the integral [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] is defined for all such values of [latex]t[\/latex]. Thus, it makes sense to consider the values of [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] as [latex]t[\/latex] approaches [latex]b[\/latex] for [latex]a<t<b[\/latex]. That is, we define [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx=\\underset{t\\to {b}^{-}}{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex], provided this limit exists. Figure 4 illustrates [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] as areas of regions for values of [latex]t[\/latex] approaching [latex]b[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_004\"><figcaption><\/figcaption><div style=\"width: 985px\" 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\/11233856\/CNX_Calc_Figure_07_07_004.jpg\" alt=\"This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and increasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. There is also a vertical asymptote at x = b. The region in the first curve is small, and progressively gets wider under the second and third graph as t gets further from a, and closer to b on the x-axis.\" width=\"975\" height=\"239\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. As [latex]t[\/latex] approaches b from the left, the value of the area from a to [latex]t[\/latex] approaches the area from a to b.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165043094066\">We use a similar approach to define [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex], where [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left(a,b\\right][\/latex] and discontinuous at [latex]a[\/latex]. We now proceed with a formal definition.<\/p>\n<div id=\"fs-id1165042333401\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<ol id=\"fs-id1165042333405\" type=\"1\">\n<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over [latex]\\left[a,b\\right)[\/latex]. Then,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042617693\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx=\\underset{t\\to {b}^{-}}{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Let [latex]f\\left(x\\right)[\/latex] be continuous over [latex]\\left(a,b\\right][\/latex]. Then,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042506827\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx=\\underset{t\\to {a}^{+}}{\\text{lim}}{\\displaystyle\\int }_{t}^{b}f\\left(x\\right)dx[\/latex].<\/div>\n<p>In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.<\/li>\n<li>If [latex]f\\left(x\\right)[\/latex] is continuous over [latex]\\left[a,b\\right][\/latex] except at a point [latex]c[\/latex] in [latex]\\left(a,b\\right)[\/latex], then<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165043378015\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx={\\displaystyle\\int }_{a}^{c}f\\left(x\\right)dx+{\\displaystyle\\int }_{c}^{b}f\\left(x\\right)dx[\/latex],<\/div>\n<p>provided both [latex]{\\displaystyle\\int }_{a}^{c}f\\left(x\\right)dx[\/latex] and [latex]{\\displaystyle\\int }_{c}^{b}f\\left(x\\right)dx[\/latex] converge. If either of these integrals diverges, then [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex] diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165043257647\">The following examples demonstrate the application of this definition.<\/p>\n<div id=\"fs-id1165043257650\" data-type=\"example\">\n<div id=\"fs-id1165043257652\" data-type=\"exercise\">\n<div id=\"fs-id1165043257655\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating a Discontinuous Integrand<\/h3>\n<div id=\"fs-id1165043257655\" data-type=\"problem\">\n<p id=\"fs-id1165043119755\">Evaluate [latex]{\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx[\/latex], if possible. State whether the integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042364402\" data-type=\"solution\">\n<p id=\"fs-id1165042364404\">The function [latex]f\\left(x\\right)=\\frac{1}{\\sqrt{4-x}}[\/latex] is continuous over [latex]\\left[0,4\\right)[\/latex] and discontinuous at 4. Using equation 1 from the definition, rewrite [latex]{\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx[\/latex] as a limit:<\/p>\n<div id=\"fs-id1165043373976\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx& =\\underset{t\\to {4}^{-}}{\\text{lim}}{\\displaystyle\\int }_{0}^{t}\\frac{1}{\\sqrt{4-x}}dx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to {4}^{-}}{\\text{lim}}\\left(-2\\sqrt{4-x}\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to {4}^{-}}{\\text{lim}}\\left(-2\\sqrt{4-t}+4\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =4.\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043173672\">The improper integral converges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043173678\" data-type=\"example\">\n<div id=\"fs-id1165043173681\" data-type=\"exercise\">\n<div id=\"fs-id1165042375631\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating a Discontinuous Integrand<\/h3>\n<div id=\"fs-id1165042375631\" data-type=\"problem\">\n<p id=\"fs-id1165042375636\">Evaluate [latex]{\\displaystyle\\int }_{0}^{2}x\\text{ln}xdx[\/latex]. State whether the integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042538864\" data-type=\"solution\">\n<p id=\"fs-id1165042538867\">Since [latex]f\\left(x\\right)=x\\ln{x}[\/latex] is continuous over [latex]\\left(0,2\\right][\/latex] and is discontinuous at zero, we can rewrite the integral in limit form using equation 2 from the definition:<\/p>\n<div id=\"fs-id1165042865839\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{2}x\\text{ln}xdx& =\\underset{t\\to {0}^{+}}{\\text{lim}}{\\displaystyle\\int }_{t}^{2}x\\text{ln}xdx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to {0}^{+}}{\\text{lim}}\\left(\\frac{1}{2}{x}^{2}\\text{ln}x-\\frac{1}{4}{x}^{2}\\right)|{}_{\\begin{array}{c}\\\\ t\\end{array}}^{\\begin{array}{c}2\\\\ \\end{array}}\\hfill & & & \\begin{array}{c}\\text{Evaluate}{\\displaystyle\\int}x\\ln{x}dx\\text{ using integration by parts}\\hfill \\\\ \\text{with }u=\\text{ln}x\\text{ and }dv=x.\\hfill \\end{array}\\hfill \\\\ & =\\underset{t\\to {0}^{+}}{\\text{lim}}\\left(2\\text{ln}2 - 1-\\frac{1}{2}{t}^{2}\\text{ln}t+\\frac{1}{4}{t}^{2}\\right).\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =2\\text{ln}2 - 1.\\hfill & & & \\begin{array}{c}\\text{Evaluate the limit.}\\underset{t\\to {0}^{+}}{\\text{lim}}{t}^{2}\\ln{t}\\text{ is indeterminate.}\\hfill \\\\ \\text{To evaluate it, rewrite as a quotient and apply}\\hfill \\\\ \\text{L'h}\\hat{o}\\text{pital's rule.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043423544\">The improper integral converges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043423550\" data-type=\"example\">\n<div id=\"fs-id1165043423552\" data-type=\"exercise\">\n<div id=\"fs-id1165042396199\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating a Discontinuous Integrand<\/h3>\n<div id=\"fs-id1165042396199\" data-type=\"problem\">\n<p id=\"fs-id1165042396204\">Evaluate [latex]{\\displaystyle\\int }_{-1}^{1}\\frac{1}{{x}^{3}}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042321413\" data-type=\"solution\">\n<p id=\"fs-id1165042321415\">Since [latex]f\\left(x\\right)=\\frac{1}{{x}^{3}}[\/latex] is discontinuous at zero, using equation 3 from the definition, we can write<\/p>\n<div id=\"fs-id1165042713598\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{-1}^{1}\\frac{1}{{x}^{3}}dx={\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx+{\\displaystyle\\int }_{0}^{1}\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042985556\">If either of the two integrals diverges, then the original integral diverges. Begin with [latex]{\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx:[\/latex]<\/p>\n<div id=\"fs-id1165042645708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx& =\\underset{t\\to {0}^{-}}{\\text{lim}}{\\displaystyle\\int }_{-1}^{t}\\frac{1}{{x}^{3}}dx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to {0}^{-}}{\\text{lim}}\\left(-\\frac{1}{2{x}^{2}}\\right)|{}_{\\begin{array}{c}\\\\ -1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to {0}^{-}}{\\text{lim}}\\left(-\\frac{1}{2{t}^{2}}+\\frac{1}{2}\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\text{+}\\infty .\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043430752\">Therefore, [latex]{\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx[\/latex] diverges. Since [latex]{\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx[\/latex] diverges, [latex]{\\displaystyle\\int }_{-1}^{1}\\frac{1}{{x}^{3}}dx[\/latex] diverges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042376188\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042376192\" data-type=\"exercise\">\n<div id=\"fs-id1165042376194\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165042376194\" data-type=\"problem\">\n<p id=\"fs-id1165042376196\">Evaluate [latex]{\\displaystyle\\int }_{0}^{2}\\frac{1}{x}dx[\/latex]. State whether the integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Hint<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042422655\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165042708426\">Write [latex]{\\displaystyle\\int }_{0}^{2}\\frac{1}{x}dx[\/latex] in limit form using equation 2 from the definition.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Show Solution<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042422640\" data-type=\"solution\">\n<p id=\"fs-id1165042422642\">[latex]+\\infty[\/latex], diverges<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8QzQO748O58?controls=0&amp;start=1388&amp;end=1468&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">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+II\/Transcripts\/3.7ImproperIntegrals1388to1468_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.7 Improper Integrals&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section id=\"fs-id1165043273047\" data-depth=\"1\"><\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1612\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.7 Improper Integrals. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":25,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.7 Improper Integrals\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1612","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1612","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1612\/revisions"}],"predecessor-version":[{"id":2720,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1612\/revisions\/2720"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/158"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1612\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1612"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1612"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1612"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1612"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}