{"id":821,"date":"2021-08-27T20:06:51","date_gmt":"2021-08-27T20:06:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=821"},"modified":"2022-10-29T00:37:32","modified_gmt":"2022-10-29T00:37:32","slug":"integrals-of-vector-valued-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/integrals-of-vector-valued-functions\/","title":{"raw":"Integrals of Vector-Valued Functions","rendered":"Integrals of Vector-Valued Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the definite integral of a vector-valued function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169739260980\" class=\" \">We introduced antiderivatives of real-valued functions in Antiderivatives and definite integrals of real-valued functions in The Definite Integral. Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the same manner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.<\/p>\r\n<p id=\"fs-id1169736616558\" class=\" \">The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at time <em data-effect=\"italics\">t<\/em>, then its antiderivative represents position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.<\/p>\r\n\r\n<div id=\"fs-id1169736616570\" class=\"ui-has-child-title\" data-type=\"note\">\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\nLet [latex]f[\/latex], [latex]g[\/latex], and [latex]h[\/latex] be integrable real-valued functions over the closed interval [latex][a,\\ b][\/latex].\r\n<ol>\r\n \t<li>The\u00a0<strong>indefinite integral of a vector-valued function<\/strong> [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}[\/latex] is\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}]\\,dt=\\Bigg[\\displaystyle\\int_{} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}[\/latex].<\/div>\r\n<div>The\u00a0<strong>definite integral of a vector-valued function<\/strong> is<\/div>\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{a}^{b} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}]\\,dt=\\Bigg[\\displaystyle\\int_{a}^{b} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{a}^{b} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}[\/latex]<\/div><\/li>\r\n \t<li>The indefinite integral of a vector-valued function [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex] is\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}]\\,dt=\\Bigg[\\displaystyle\\int_{} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}+\\Bigg[\\displaystyle\\int_{} \\ h\\,(t)\\,dt\\Bigg]\\,{\\bf{k}}[\/latex]<\/div>\r\n<div>The definite integral of the vector-valued function is<\/div>\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{a}^{b} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}]\\,dt=\\Bigg[\\displaystyle\\int_{a}^{b} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{a}^{b} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}+\\Bigg[\\displaystyle\\int_{a}^{b} \\ h\\,(t)\\,dt\\Bigg]\\,{\\bf{k}}.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\nSince the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each of these component integrals contains an integration constant. They can all be different. For example, in the two-dimensional case, we can have\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ f\\,(t)\\,dt=F\\,(t)+C_{1}[\/latex] and [latex]\\displaystyle\\int_{} \\ g\\,(t)\\,dt=G\\,(t)+C_{2}[\/latex],<\/p>\r\nwhere F and G are antiderivatives of [latex]f[\/latex] and [latex]g[\/latex], respectively. Then\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int_{} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}]\\,dt} &amp; =\\hfill &amp; {\\Bigg[\\displaystyle\\int_{} \\ f\\,(t)\\,dt\\Bigg]\\,{\\bf{i}}+\\Bigg[\\int_{} \\ g\\,(t)\\,dt\\Bigg]\\,{\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(F\\,(t)+C_{1})\\,{\\bf{i}}+(G\\,(t)+C_{2})\\,{\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {F\\,(t)\\,{\\bf{i}}+G\\,(t)\\,{\\bf{j}}+C_{1}\\,{\\bf{i}}+C_{2}\\,{\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {F\\,(t)\\,{\\bf{i}}+G\\,(t)\\,{\\bf{j}}+C,} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nwhere [latex]C=C_{1}\\,{\\bf{i}}+C_{2}\\,{\\bf{j}}[\/latex]. Therefore, the integration constant becomes a constant vector.\r\n\r\nSince we will also encounter integrals frequently throughout this course, we review several common integrals below.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Integrals of common functions<\/h3>\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li>[latex] \\int x^n \\ dx = \\frac{1}{n+1}x^{n+1} + C \\ (\\text{ if } x \\ne -1) [\/latex]<\/li>\r\n \t<li>[latex] \\int \\frac{1}{x} \\ dx = \\ln |x|+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int e^u \\ du = e^u+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int \\cos u \\ du = \\sin u+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int \\sin u \\ du = - \\cos u+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int \\sec u \\tan u \\ du = \\sec u+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int \\csc u \\cot u \\ du = -\\csc u+ C[\/latex]<\/li>\r\n \t<li>[latex] \\int \\sec^2 u \\ du = \\tan u+ C[\/latex]<\/li>\r\n \t<li>[latex] \\int \\csc^2 u \\ du = -\\cot u+ C [\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li>[latex] \\int \\tan u \\ du = \\ln |\\sec u|+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int \\sec u \\ du = \\ln |\\sec u + \\tan u]+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int \\frac{1}{a^2+u^2} \\ du = \\frac{1}{a}\\arctan \\left( \\frac{u}{a} \\right)+ C [\/latex]<\/li>\r\n \t<li>[latex] \\int \\frac{1}{\\sqrt{a^2-u^2}} \\ du = \\arcsin \\left( \\frac{u}{a} \\right)+ C [\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0integrating vector-valued functions<\/h3>\r\nCalculate each of the following integrals:\r\n<ol>\r\n \t<li style=\"text-align: left;\">[latex]\\displaystyle\\int_{} \\big[(3t^{2}+2t)\\,{\\bf{i}}+(3t-6)\\,{\\bf{j}}+(6t^{3}+5t^{2}-4)\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int_{} \\big[{\\langle}t,\\ t^{2},\\ t^{3}\\rangle\\times{\\langle}t^{3},\\ t^{2},\\ t\\rangle\\big]\\,dt[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int_{0}^{\\frac{\\pi}{3}}\\big[\\sin{2t\\,{\\bf{i}}}+\\tan{t\\,{\\bf{j}}}+e^{-2t}\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055198\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055198\"]\r\n\r\na. \u00a0We use the first part of the definition of the integral of a space curve:\r\n<div class=\"exercise\">\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ \\big[(3t^{2}+2t)\\,{\\bf{i}}+(3t-6)\\,{\\bf{j}}+(6t^{3}+5t^{2}-4)\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill &amp; =\\hfill &amp; {\\Big[\\displaystyle\\int_{} \\ 3t^{2}+2t\\,dt\\Big]\\,{\\bf{i}}+\\Big[\\displaystyle\\int_{} \\ 3t-6\\,dt\\Big]\\,{\\bf{j}}+\\Big[\\displaystyle\\int_{} 6t^{3}+5t^{2}-4\\,dt\\Big]\\,{\\bf{k}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(t^{3}+t^{2})\\,{\\bf{i}}+(\\frac{3}{2}t^{2}-6t)\\,{\\bf{j}}+(\\frac{3}{2}t^{4}+\\frac{5}{3}t^{3}-4t)\\,{\\bf{k}}+C.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p style=\"text-align: left;\">b. \u00a0First calculate [latex]\\langle{t},\\ t^{2},\\ t^{3}\\rangle\\times\\langle{t}^{3},\\ t^{2},\\ {t}\\rangle[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\langle{t},\\ t^{2},\\ t^{3}\\rangle\\times\\langle{t}^{3},\\ t^{2},\\ t\\rangle} &amp; =\\hfill &amp; {\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}}\\\\t&amp;t^{2}&amp;t^{3}\\\\t^{3}&amp;t^{2}&amp;t\\end{vmatrix}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\big(t^{2}\\,(t)-t^{3}\\,(t^{2})\\big)\\,{\\bf{i}}-\\big(t^{2}-t^{3}\\,(t^{3})\\big)\\,{\\bf{j}}+\\big(t\\,(t^{2})-t^{2}\\,(t^{3})\\big)\\,{\\bf{k}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(t^{3}-t^{5})\\,{\\bf{i}}+(t^{6}-t^{2})\\,{\\bf{j}}+(t^{3}-t^{5})\\,{\\bf{k}}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, substitute this back into the integral and integrate:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int_{} \\ \\big[\\langle{t},\\ t^{2},\\ t^{3}\\rangle\\times\\langle{t}^{3},\\ t^{2},\\ t\\rangle\\big]\\,dt} &amp; =\\hfill &amp; {\\displaystyle\\int_{} \\ (t^{3}-t^{5})\\,{\\bf{i}}+(t^{6}-t^{2})\\,{\\bf{j}}+(t^{3}-t^{5})\\,{\\bf{k}}\\,dt} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\big(\\frac{t^{4}}{4}-\\frac{t^{6}}{6}\\big)\\,{\\bf{i}}+\\big(\\frac{t^{7}}{7}-\\frac{t^{3}}{3}\\big)\\,{\\bf{j}}+\\big(\\frac{t^{4}}{4}-\\frac{t^{6}}{6}\\big)\\,{\\bf{k}}+C.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nc. \u00a0Use the second part of the definition of the integral of a space curve:\r\n<div style=\"text-align: left;\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [latex]\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\big[\\sin{2t\\,{\\bf{i}}}+\\tan{t\\,{\\bf{j}}}+e^{-2t}\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill &amp; =\\hfill &amp; {\\Big[\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\sin{2t}\\,dt\\Big]\\,{\\bf{i}}+\\Big[\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\tan{t}\\,dt\\Big]\\,{\\bf{j}}+\\Big[\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\ e^{-2t}\\,dt\\Big]{\\bf{k}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\big(-\\frac{1}{2}\\cos{2t}\\big)\\Big{|}_0^{\\frac{\\pi}{3}}\\,{\\bf{i}}-(\\ln{(\\cos{t})})\\Big{|}_0^{\\frac{\\pi}{3}}\\,{\\bf{j}}-\\big(\\frac{1}{2}e^{-2t}\\big)\\Big{|}_0^{\\frac{\\pi}{3}}\\,{\\bf{k}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\big(-\\frac{1}{2}\\cos{\\frac{2\\pi}{3}}+\\frac{1}{2}\\cos{0})\\big)\\,{\\bf{i}}-\\big(\\ln{(\\cos{\\frac{\\pi}{3})}}-\\ln{(\\cos{0}}\\big)\\,{\\bf{j}}-\\big(\\frac{1}{2}\\,e^{-2\\pi\/3}-\\frac{1}{2}\\,e^{-2(0)}\\big)\\,{\\bf{k}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\big(\\frac{1}{4}+\\frac{1}{2}\\big)\\,{\\bf{i}}-(-\\ln{2})\\,{\\bf{j}}-\\big(\\frac{1}{2}\\,e^{-2\\pi\/3}-\\frac{1}{2}\\big)\\,{\\bf{k}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{3}{4}\\,{\\bf{i}}+(\\ln{2})\\,{\\bf{j}}+\\big(\\frac{1}{2}-\\frac{1}{2}\\,e^{-2\\pi\/3}\\big)\\,{\\bf{k}}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nCalculate the following integral:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055279\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055279\"]\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt=16\\,{\\bf{i}}+10\\,{\\bf{j}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\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 src=\"\/\/plugin.3playmedia.com\/show?mf=7949613&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VeLUdurbQN4&amp;video_target=tpm-plugin-4s0tfrxc-VeLUdurbQN4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.8_transcript.html\">transcript for \u201cCP 3.8\u201d here (opens in new window).<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the definite integral of a vector-valued function<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169739260980\" class=\"\">We introduced antiderivatives of real-valued functions in Antiderivatives and definite integrals of real-valued functions in The Definite Integral. Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the same manner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.<\/p>\n<p id=\"fs-id1169736616558\" class=\"\">The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at time <em data-effect=\"italics\">t<\/em>, then its antiderivative represents position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.<\/p>\n<div id=\"fs-id1169736616570\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]f[\/latex], [latex]g[\/latex], and [latex]h[\/latex] be integrable real-valued functions over the closed interval [latex][a,\\ b][\/latex].<\/p>\n<ol>\n<li>The\u00a0<strong>indefinite integral of a vector-valued function<\/strong> [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}[\/latex] is\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}]\\,dt=\\Bigg[\\displaystyle\\int_{} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}[\/latex].<\/div>\n<div>The\u00a0<strong>definite integral of a vector-valued function<\/strong> is<\/div>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{a}^{b} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}]\\,dt=\\Bigg[\\displaystyle\\int_{a}^{b} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{a}^{b} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}[\/latex]<\/div>\n<\/li>\n<li>The indefinite integral of a vector-valued function [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex] is\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}]\\,dt=\\Bigg[\\displaystyle\\int_{} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}+\\Bigg[\\displaystyle\\int_{} \\ h\\,(t)\\,dt\\Bigg]\\,{\\bf{k}}[\/latex]<\/div>\n<div>The definite integral of the vector-valued function is<\/div>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{a}^{b} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}]\\,dt=\\Bigg[\\displaystyle\\int_{a}^{b} \\ f\\,(t)\\,dt\\Bigg]{\\bf{i}}+\\Bigg[\\displaystyle\\int_{a}^{b} \\ g\\,(t)\\,dt\\Bigg]{\\bf{j}}+\\Bigg[\\displaystyle\\int_{a}^{b} \\ h\\,(t)\\,dt\\Bigg]\\,{\\bf{k}}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p>Since the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each of these component integrals contains an integration constant. They can all be different. For example, in the two-dimensional case, we can have<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ f\\,(t)\\,dt=F\\,(t)+C_{1}[\/latex] and [latex]\\displaystyle\\int_{} \\ g\\,(t)\\,dt=G\\,(t)+C_{2}[\/latex],<\/p>\n<p>where F and G are antiderivatives of [latex]f[\/latex] and [latex]g[\/latex], respectively. Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int_{} \\ [f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}]\\,dt} & =\\hfill & {\\Bigg[\\displaystyle\\int_{} \\ f\\,(t)\\,dt\\Bigg]\\,{\\bf{i}}+\\Bigg[\\int_{} \\ g\\,(t)\\,dt\\Bigg]\\,{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {(F\\,(t)+C_{1})\\,{\\bf{i}}+(G\\,(t)+C_{2})\\,{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {F\\,(t)\\,{\\bf{i}}+G\\,(t)\\,{\\bf{j}}+C_{1}\\,{\\bf{i}}+C_{2}\\,{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {F\\,(t)\\,{\\bf{i}}+G\\,(t)\\,{\\bf{j}}+C,} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>where [latex]C=C_{1}\\,{\\bf{i}}+C_{2}\\,{\\bf{j}}[\/latex]. Therefore, the integration constant becomes a constant vector.<\/p>\n<p>Since we will also encounter integrals frequently throughout this course, we review several common integrals below.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Integrals of common functions<\/h3>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li>[latex]\\int x^n \\ dx = \\frac{1}{n+1}x^{n+1} + C \\ (\\text{ if } x \\ne -1)[\/latex]<\/li>\n<li>[latex]\\int \\frac{1}{x} \\ dx = \\ln |x|+ C[\/latex]<\/li>\n<li>[latex]\\int e^u \\ du = e^u+ C[\/latex]<\/li>\n<li>[latex]\\int \\cos u \\ du = \\sin u+ C[\/latex]<\/li>\n<li>[latex]\\int \\sin u \\ du = - \\cos u+ C[\/latex]<\/li>\n<li>[latex]\\int \\sec u \\tan u \\ du = \\sec u+ C[\/latex]<\/li>\n<li>[latex]\\int \\csc u \\cot u \\ du = -\\csc u+ C[\/latex]<\/li>\n<li>[latex]\\int \\sec^2 u \\ du = \\tan u+ C[\/latex]<\/li>\n<li>[latex]\\int \\csc^2 u \\ du = -\\cot u+ C[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li>[latex]\\int \\tan u \\ du = \\ln |\\sec u|+ C[\/latex]<\/li>\n<li>[latex]\\int \\sec u \\ du = \\ln |\\sec u + \\tan u]+ C[\/latex]<\/li>\n<li>[latex]\\int \\frac{1}{a^2+u^2} \\ du = \\frac{1}{a}\\arctan \\left( \\frac{u}{a} \\right)+ C[\/latex]<\/li>\n<li>[latex]\\int \\frac{1}{\\sqrt{a^2-u^2}} \\ du = \\arcsin \\left( \\frac{u}{a} \\right)+ C[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0integrating vector-valued functions<\/h3>\n<p>Calculate each of the following integrals:<\/p>\n<ol>\n<li style=\"text-align: left;\">[latex]\\displaystyle\\int_{} \\big[(3t^{2}+2t)\\,{\\bf{i}}+(3t-6)\\,{\\bf{j}}+(6t^{3}+5t^{2}-4)\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int_{} \\big[{\\langle}t,\\ t^{2},\\ t^{3}\\rangle\\times{\\langle}t^{3},\\ t^{2},\\ t\\rangle\\big]\\,dt[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int_{0}^{\\frac{\\pi}{3}}\\big[\\sin{2t\\,{\\bf{i}}}+\\tan{t\\,{\\bf{j}}}+e^{-2t}\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055198\">Show Solution<\/span><\/div>\n<div id=\"qfs-id1167794055198\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. \u00a0We use the first part of the definition of the integral of a space curve:<\/p>\n<div class=\"exercise\">\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{} \\ \\big[(3t^{2}+2t)\\,{\\bf{i}}+(3t-6)\\,{\\bf{j}}+(6t^{3}+5t^{2}-4)\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill & =\\hfill & {\\Big[\\displaystyle\\int_{} \\ 3t^{2}+2t\\,dt\\Big]\\,{\\bf{i}}+\\Big[\\displaystyle\\int_{} \\ 3t-6\\,dt\\Big]\\,{\\bf{j}}+\\Big[\\displaystyle\\int_{} 6t^{3}+5t^{2}-4\\,dt\\Big]\\,{\\bf{k}}} \\hfill \\\\ \\hfill & =\\hfill & {(t^{3}+t^{2})\\,{\\bf{i}}+(\\frac{3}{2}t^{2}-6t)\\,{\\bf{j}}+(\\frac{3}{2}t^{4}+\\frac{5}{3}t^{3}-4t)\\,{\\bf{k}}+C.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\">b. \u00a0First calculate [latex]\\langle{t},\\ t^{2},\\ t^{3}\\rangle\\times\\langle{t}^{3},\\ t^{2},\\ {t}\\rangle[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\langle{t},\\ t^{2},\\ t^{3}\\rangle\\times\\langle{t}^{3},\\ t^{2},\\ t\\rangle} & =\\hfill & {\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}}\\\\t&t^{2}&t^{3}\\\\t^{3}&t^{2}&t\\end{vmatrix}} \\hfill \\\\ \\hfill & =\\hfill & {\\big(t^{2}\\,(t)-t^{3}\\,(t^{2})\\big)\\,{\\bf{i}}-\\big(t^{2}-t^{3}\\,(t^{3})\\big)\\,{\\bf{j}}+\\big(t\\,(t^{2})-t^{2}\\,(t^{3})\\big)\\,{\\bf{k}}} \\hfill \\\\ \\hfill & =\\hfill & {(t^{3}-t^{5})\\,{\\bf{i}}+(t^{6}-t^{2})\\,{\\bf{j}}+(t^{3}-t^{5})\\,{\\bf{k}}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, substitute this back into the integral and integrate:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int_{} \\ \\big[\\langle{t},\\ t^{2},\\ t^{3}\\rangle\\times\\langle{t}^{3},\\ t^{2},\\ t\\rangle\\big]\\,dt} & =\\hfill & {\\displaystyle\\int_{} \\ (t^{3}-t^{5})\\,{\\bf{i}}+(t^{6}-t^{2})\\,{\\bf{j}}+(t^{3}-t^{5})\\,{\\bf{k}}\\,dt} \\hfill \\\\ \\hfill & =\\hfill & {\\big(\\frac{t^{4}}{4}-\\frac{t^{6}}{6}\\big)\\,{\\bf{i}}+\\big(\\frac{t^{7}}{7}-\\frac{t^{3}}{3}\\big)\\,{\\bf{j}}+\\big(\\frac{t^{4}}{4}-\\frac{t^{6}}{6}\\big)\\,{\\bf{k}}+C.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>c. \u00a0Use the second part of the definition of the integral of a space curve:<\/p>\n<div style=\"text-align: left;\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [latex]\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\big[\\sin{2t\\,{\\bf{i}}}+\\tan{t\\,{\\bf{j}}}+e^{-2t}\\,{\\bf{k}}\\big]\\,dt[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill & =\\hfill & {\\Big[\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\sin{2t}\\,dt\\Big]\\,{\\bf{i}}+\\Big[\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\tan{t}\\,dt\\Big]\\,{\\bf{j}}+\\Big[\\displaystyle\\int_{0}^{\\frac{\\pi}{3}} \\ e^{-2t}\\,dt\\Big]{\\bf{k}}} \\hfill \\\\ \\hfill & =\\hfill & {\\big(-\\frac{1}{2}\\cos{2t}\\big)\\Big{|}_0^{\\frac{\\pi}{3}}\\,{\\bf{i}}-(\\ln{(\\cos{t})})\\Big{|}_0^{\\frac{\\pi}{3}}\\,{\\bf{j}}-\\big(\\frac{1}{2}e^{-2t}\\big)\\Big{|}_0^{\\frac{\\pi}{3}}\\,{\\bf{k}}} \\hfill \\\\ \\hfill & =\\hfill & {\\big(-\\frac{1}{2}\\cos{\\frac{2\\pi}{3}}+\\frac{1}{2}\\cos{0})\\big)\\,{\\bf{i}}-\\big(\\ln{(\\cos{\\frac{\\pi}{3})}}-\\ln{(\\cos{0}}\\big)\\,{\\bf{j}}-\\big(\\frac{1}{2}\\,e^{-2\\pi\/3}-\\frac{1}{2}\\,e^{-2(0)}\\big)\\,{\\bf{k}}} \\hfill \\\\ \\hfill & =\\hfill & {\\big(\\frac{1}{4}+\\frac{1}{2}\\big)\\,{\\bf{i}}-(-\\ln{2})\\,{\\bf{j}}-\\big(\\frac{1}{2}\\,e^{-2\\pi\/3}-\\frac{1}{2}\\big)\\,{\\bf{k}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{3}{4}\\,{\\bf{i}}+(\\ln{2})\\,{\\bf{j}}+\\big(\\frac{1}{2}-\\frac{1}{2}\\,e^{-2\\pi\/3}\\big)\\,{\\bf{k}}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Calculate the following integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt[\/latex]<\/p>\n<div id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055279\">Show Solution<\/span><\/div>\n<div id=\"qfs-id1167794055279\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt=16\\,{\\bf{i}}+10\\,{\\bf{j}}[\/latex]<\/p>\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\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949613&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VeLUdurbQN4&amp;video_target=tpm-plugin-4s0tfrxc-VeLUdurbQN4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.8_transcript.html\">transcript for \u201cCP 3.8\u201d here (opens in new window).<\/a><\/div>\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-821\">\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>CP 3.8. <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 3. <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-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/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-3\/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":349141,"menu_order":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) 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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-821","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/821","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":190,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/821\/revisions"}],"predecessor-version":[{"id":4806,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/821\/revisions\/4806"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/821\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=821"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=821"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=821"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}