{"id":4091,"date":"2022-04-14T18:15:53","date_gmt":"2022-04-14T18:15:53","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-arc-length-and-curvature\/"},"modified":"2022-04-18T16:24:52","modified_gmt":"2022-04-18T16:24:52","slug":"skills-review-for-arc-length-and-curvature","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-arc-length-and-curvature\/","title":{"raw":"Skills Review for Arc Length and Curvature","rendered":"Skills Review for Arc Length and Curvature"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals<\/li>\r\n \t<li>Use a table of integrals to solve integration problems<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Arc Length and Curvature section, we will explore the curvature of vector-valued functions. Here we will review how to apply the Second Fundamental Theorem of Calculus and use integration table formulas.\r\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\r\n<p id=\"fs-id1170572622222\">The <strong>Fundamental Theorem of Calculus, Part 2<\/strong> (also known as the <strong>evaluation theorem<\/strong>) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.<\/p>\r\n\r\n<div id=\"fs-id1170571660076\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\r\n\r\n<hr>\r\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572430375\" class=\"bc-section section\">\r\n<div id=\"fs-id1170571678911\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating an Integral with the Fundamental Theorem of Calculus<\/h3>\r\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\r\n\r\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\r\n<div id=\"fs-id1170571678913\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1170572173704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572173704\"]\r\n<p id=\"fs-id1170572173704\">Recall the power rule for <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/antiderivatives\/\">Antiderivatives<\/a>:<\/p>\r\n\r\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\r\n\r\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill &amp; =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ &amp; =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ &amp; =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ &amp; =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ &amp; =\\frac{16}{3}-16\\hfill \\\\ &amp; =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571561288\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2<\/h3>\r\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572130389\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572130389\"]\r\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\r\n\r\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\r\n\r\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\r\n\r\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill &amp; ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ &amp; =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ &amp; =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ &amp; =18-6-\\frac{2}{3}+2\\hfill \\\\ &amp; =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571609442\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]\r\n\r\n[reveal-answer q=\"3887125\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3887125\"]\r\n\r\nUse the power rule.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571639103\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571639103\"]\r\n\r\n[latex]\\frac{7}{24}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]211336[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Tables of Integrals<\/h2>\r\n<p id=\"fs-id1165042105278\">Integration tables, if used in the right manner, can be a handy way either to evaluate or check an integral quickly. Keep in mind that when using a table to check an answer, it is possible for two completely correct solutions to look very different. For example, in Trigonometric Substitution, we found that, by using the substitution [latex]x=\\tan\\theta [\/latex], we can arrive at<\/p>\r\n\r\n<div id=\"fs-id1165040681970\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)+C[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041951596\">However, using [latex]x=\\text{sinh}\\theta [\/latex], we obtained a different solution\u2014namely,<\/p>\r\n\r\n<div id=\"fs-id1165041848129\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}={\\text{sinh}}^{-1}x+C[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041948411\">We later showed algebraically that the two solutions are equivalent. That is, we showed that [latex]{\\text{sinh}}^{-1}x=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex]. In this case, the two antiderivatives that we found were actually equal. This need not be the case. However, as long as the difference in the two antiderivatives is a constant, they are equivalent.<\/p>\r\n\r\n<div id=\"fs-id1165041930726\" data-type=\"example\">\r\n<div id=\"fs-id1165041787566\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042085684\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:&nbsp;Using a Formula from a Table to Evaluate an Integral<\/h3>\r\n<div id=\"fs-id1165042085684\" data-type=\"problem\">\r\n<p id=\"fs-id1165041973418\">Use the table formula<\/p>\r\n\r\n<div id=\"fs-id1165041758432\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040771022\">to evaluate [latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx[\/latex].<\/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-id1165041913720\" data-type=\"solution\">\r\n<p id=\"fs-id1165041766265\">If we look at integration tables, we see that several formulas contain expressions of the form [latex]\\sqrt{{a}^{2}-{u}^{2}}[\/latex]. This expression is actually similar to [latex]\\sqrt{16-{e}^{2x}}[\/latex], where [latex]a=4[\/latex] and [latex]u={e}^{x}[\/latex]. Keep in mind that we must also have [latex]du={e}^{x}[\/latex]. Multiplying the numerator and the denominator of the given integral by [latex]{e}^{x}[\/latex] should help to put this integral in a useful form. Thus, we now have<\/p>\r\n\r\n<div id=\"fs-id1165040752681\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx=\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041797345\">Substituting [latex]u={e}^{x}[\/latex] and [latex]du={e}^{x}[\/latex] produces [latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du[\/latex]. From the integration table (#88 in Appendix A),<\/p>\r\n\r\n<div id=\"fs-id1165042045994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041845825\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165040798665\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx}&amp; ={\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx}\\hfill &amp; &amp; &amp; \\text{Substitute}u={e}^{x}\\text{and}du={e}^{x}dx.\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{\\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du}\\hfill &amp; &amp; &amp; \\text{Apply the formula using}a=4.\\hfill \\\\ &amp; =-\\frac{\\sqrt{{4}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{4}+C\\hfill &amp; &amp; &amp; \\text{Substitute}u={e}^{x}.\\hfill \\\\ &amp; =-\\frac{\\sqrt{16-{e}^{2x}}}{u}-{\\sin}^{-1}\\left(\\frac{{e}^{x}}{4}\\right)+C.\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe full integration table with formulas can be found in Appendix A.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals<\/li>\n<li>Use a table of integrals to solve integration problems<\/li>\n<\/ul>\n<\/div>\n<p>In the Arc Length and Curvature section, we will explore the curvature of vector-valued functions. Here we will review how to apply the Second Fundamental Theorem of Calculus and use integration table formulas.<\/p>\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\n<p id=\"fs-id1170572622222\">The <strong>Fundamental Theorem of Calculus, Part 2<\/strong> (also known as the <strong>evaluation theorem<\/strong>) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.<\/p>\n<div id=\"fs-id1170571660076\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\n<hr \/>\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then<\/p>\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1170572430375\" class=\"bc-section section\">\n<div id=\"fs-id1170571678911\" class=\"textbook exercises\">\n<h3>Example: Evaluating an Integral with the Fundamental Theorem of Calculus<\/h3>\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\n<div id=\"fs-id1170571678913\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572173704\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572173704\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572173704\">Recall the power rule for <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/antiderivatives\/\">Antiderivatives<\/a>:<\/p>\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill & =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ & =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ & =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ & =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ & =\\frac{16}{3}-16\\hfill \\\\ & =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571561288\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2<\/h3>\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572130389\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572130389\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill & ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ & =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ & =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ & =18-6-\\frac{2}{3}+2\\hfill \\\\ & =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571609442\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3887125\">Hint<\/span><\/p>\n<div id=\"q3887125\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the power rule.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571639103\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571639103\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{7}{24}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm211336\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=211336&theme=oea&iframe_resize_id=ohm211336&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Tables of Integrals<\/h2>\n<p id=\"fs-id1165042105278\">Integration tables, if used in the right manner, can be a handy way either to evaluate or check an integral quickly. Keep in mind that when using a table to check an answer, it is possible for two completely correct solutions to look very different. For example, in Trigonometric Substitution, we found that, by using the substitution [latex]x=\\tan\\theta[\/latex], we can arrive at<\/p>\n<div id=\"fs-id1165040681970\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041951596\">However, using [latex]x=\\text{sinh}\\theta[\/latex], we obtained a different solution\u2014namely,<\/p>\n<div id=\"fs-id1165041848129\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}={\\text{sinh}}^{-1}x+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041948411\">We later showed algebraically that the two solutions are equivalent. That is, we showed that [latex]{\\text{sinh}}^{-1}x=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex]. In this case, the two antiderivatives that we found were actually equal. This need not be the case. However, as long as the difference in the two antiderivatives is a constant, they are equivalent.<\/p>\n<div id=\"fs-id1165041930726\" data-type=\"example\">\n<div id=\"fs-id1165041787566\" data-type=\"exercise\">\n<div id=\"fs-id1165042085684\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:&nbsp;Using a Formula from a Table to Evaluate an Integral<\/h3>\n<div id=\"fs-id1165042085684\" data-type=\"problem\">\n<p id=\"fs-id1165041973418\">Use the table formula<\/p>\n<div id=\"fs-id1165041758432\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040771022\">to evaluate [latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx[\/latex].<\/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-id1165041913720\" data-type=\"solution\">\n<p id=\"fs-id1165041766265\">If we look at integration tables, we see that several formulas contain expressions of the form [latex]\\sqrt{{a}^{2}-{u}^{2}}[\/latex]. This expression is actually similar to [latex]\\sqrt{16-{e}^{2x}}[\/latex], where [latex]a=4[\/latex] and [latex]u={e}^{x}[\/latex]. Keep in mind that we must also have [latex]du={e}^{x}[\/latex]. Multiplying the numerator and the denominator of the given integral by [latex]{e}^{x}[\/latex] should help to put this integral in a useful form. Thus, we now have<\/p>\n<div id=\"fs-id1165040752681\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx=\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041797345\">Substituting [latex]u={e}^{x}[\/latex] and [latex]du={e}^{x}[\/latex] produces [latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du[\/latex]. From the integration table (#88 in Appendix A),<\/p>\n<div id=\"fs-id1165042045994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041845825\">Thus,<\/p>\n<div id=\"fs-id1165040798665\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx}& ={\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx}\\hfill & & & \\text{Substitute}u={e}^{x}\\text{and}du={e}^{x}dx.\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du}\\hfill & & & \\text{Apply the formula using}a=4.\\hfill \\\\ & =-\\frac{\\sqrt{{4}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{4}+C\\hfill & & & \\text{Substitute}u={e}^{x}.\\hfill \\\\ & =-\\frac{\\sqrt{16-{e}^{2x}}}{u}-{\\sin}^{-1}\\left(\\frac{{e}^{x}}{4}\\right)+C.\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The full integration table with formulas can be found in Appendix A.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":349141,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4091","chapter","type-chapter","status-publish","hentry"],"part":4118,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4091","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4091\/revisions"}],"predecessor-version":[{"id":4231,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4091\/revisions\/4231"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4118"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4091\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4091"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4091"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4091"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4091"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}