{"id":956,"date":"2025-10-14T16:52:44","date_gmt":"2025-10-14T16:52:44","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=956"},"modified":"2025-10-22T16:39:23","modified_gmt":"2025-10-22T16:39:23","slug":"calculus-of-vector-valued-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/calculus-of-vector-valued-functions-learn-it-3\/","title":{"raw":"Calculus of Vector-Valued Functions: Learn It 3","rendered":"Calculus of Vector-Valued Functions: Learn It 3"},"content":{"raw":"

Integrals of Vector-Valued Functions<\/h2>\r\nYou've already worked with antiderivatives and definite integrals of real-valued functions. These concepts extend naturally to vector-valued functions using the same component-by-component approach we used for derivatives.\r\n

Just as we can differentiate a vector-valued function by differentiating each component separately, we can calculate antiderivatives and integrals the same way. The Fundamental Theorem of Calculus applies to vector-valued functions as well.<\/p>\r\n

Antiderivatives of vector-valued functions appear frequently in applications. If a vector-valued function represents the velocity of an object at time [latex]t[\/latex], then its antiderivative represents position. If the function represents acceleration, then the antiderivative represents velocity.<\/p>\r\n\r\n

\r\n

integrals of vector-valued functions<\/h3>\r\n

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>\r\n

For plane curves:<\/strong><\/p>\r\n\r\n

    \r\n \t
  • Indefinite integral:<\/strong> [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\r\n \t
  • Definite integral:<\/strong> [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\r\n<\/ul>\r\n

    For space curves:<\/strong><\/p>\r\n\r\n

      \r\n \t
    • Indefinite integral:<\/strong> [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\r\n \t
    • Definite integral:<\/strong> [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int_{a}^{b} h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
      \r\n

      When you integrate a vector-valued function, each component integral includes its own integration constant. These constants can be combined into a single constant vector [latex]{\\bf{C}}[\/latex].<\/p>\r\n

      For example, if [latex]\\int f(t)dt=F(t)+C_{1}[\/latex] and [latex]\\int g(t)dt=G(t)+C_{2}[\/latex], then:<\/p>\r\n

      [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=F(t){\\bf{i}}+G(t){\\bf{j}}+C_{1}{\\bf{i}}+C_{2}{\\bf{j}}=F(t){\\bf{i}}+G(t){\\bf{j}}+{\\bf{C}}[\/latex]<\/p>\r\n

      where [latex]{\\bf{C}}=C_{1}{\\bf{i}}+C_{2}{\\bf{j}}[\/latex].<\/p>\r\n\r\n<\/section>You'll use the same integration formulas you learned in Calculus 1 to integrate each component function. Since these formulas are essential tools for the work ahead, here's a quick reference of the most commonly used integrals.\r\n\r\n

      \r\n

      Integrals of Common Functions<\/strong><\/p>\r\n\r\n

        \r\n \t
      • [latex]\\int x^n dx = \\frac{1}{n+1}x^{n+1} + C[\/latex] (if [latex]n \\ne -1[\/latex])<\/li>\r\n \t
      • [latex]\\int \\frac{1}{x} dx = \\ln |x|+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int e^u du = e^u+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\cos u du = \\sin u+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\sin u du = -\\cos u+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\sec^2 u du = \\tan u+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\csc^2 u du = -\\cot u+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\sec u \\tan u du = \\sec u+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\csc u \\cot u du = -\\csc u+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\tan u du = \\ln |\\sec u|+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\sec u du = \\ln |\\sec u + \\tan u|+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\frac{1}{a^2+u^2} du = \\frac{1}{a}\\arctan \\left( \\frac{u}{a} \\right)+ C[\/latex]<\/li>\r\n \t
      • [latex]\\int \\frac{1}{\\sqrt{a^2-u^2}} du = \\arcsin \\left( \\frac{u}{a} \\right)+ C[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
        Calculate each of the following integrals:\r\n
          \r\n \t
        1. [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
        2. [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
        3. [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
          [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
          \r\n
          [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
          [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>\r\n \r\n

          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

          [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>\r\n \r\n\r\nNext, substitute this back into the integral and integrate:\r\n
          [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>\r\n \r\n\r\nc. \u00a0Use the second part of the definition of the integral of a space curve:\r\n
          \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
          [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>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

          Integrals of Vector-Valued Functions<\/h2>\n

          You’ve already worked with antiderivatives and definite integrals of real-valued functions. These concepts extend naturally to vector-valued functions using the same component-by-component approach we used for derivatives.<\/p>\n

          Just as we can differentiate a vector-valued function by differentiating each component separately, we can calculate antiderivatives and integrals the same way. The Fundamental Theorem of Calculus applies to vector-valued functions as well.<\/p>\n

          Antiderivatives of vector-valued functions appear frequently in applications. If a vector-valued function represents the velocity of an object at time [latex]t[\/latex], then its antiderivative represents position. If the function represents acceleration, then the antiderivative represents velocity.<\/p>\n

          \n

          integrals of vector-valued functions<\/h3>\n

          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

          For plane curves:<\/strong><\/p>\n

            \n
          • Indefinite integral:<\/strong> [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\n
          • Definite integral:<\/strong> [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\n<\/ul>\n

            For space curves:<\/strong><\/p>\n

              \n
            • Indefinite integral:<\/strong> [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\n
            • Definite integral:<\/strong> [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int_{a}^{b} h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\n<\/ul>\n<\/section>\n
              \n

              When you integrate a vector-valued function, each component integral includes its own integration constant. These constants can be combined into a single constant vector [latex]{\\bf{C}}[\/latex].<\/p>\n

              For example, if [latex]\\int f(t)dt=F(t)+C_{1}[\/latex] and [latex]\\int g(t)dt=G(t)+C_{2}[\/latex], then:<\/p>\n

              [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=F(t){\\bf{i}}+G(t){\\bf{j}}+C_{1}{\\bf{i}}+C_{2}{\\bf{j}}=F(t){\\bf{i}}+G(t){\\bf{j}}+{\\bf{C}}[\/latex]<\/p>\n

              where [latex]{\\bf{C}}=C_{1}{\\bf{i}}+C_{2}{\\bf{j}}[\/latex].<\/p>\n<\/section>\n

              You’ll use the same integration formulas you learned in Calculus 1 to integrate each component function. Since these formulas are essential tools for the work ahead, here’s a quick reference of the most commonly used integrals.<\/p>\n

              \n

              Integrals of Common Functions<\/strong><\/p>\n

                \n
              • [latex]\\int x^n dx = \\frac{1}{n+1}x^{n+1} + C[\/latex] (if [latex]n \\ne -1[\/latex])<\/li>\n
              • [latex]\\int \\frac{1}{x} dx = \\ln |x|+ C[\/latex]<\/li>\n
              • [latex]\\int e^u du = e^u+ C[\/latex]<\/li>\n
              • [latex]\\int \\cos u du = \\sin u+ C[\/latex]<\/li>\n
              • [latex]\\int \\sin u du = -\\cos u+ C[\/latex]<\/li>\n
              • [latex]\\int \\sec^2 u du = \\tan u+ C[\/latex]<\/li>\n
              • [latex]\\int \\csc^2 u du = -\\cot u+ C[\/latex]<\/li>\n
              • [latex]\\int \\sec u \\tan u du = \\sec u+ C[\/latex]<\/li>\n
              • [latex]\\int \\csc u \\cot u du = -\\csc u+ C[\/latex]<\/li>\n
              • [latex]\\int \\tan u du = \\ln |\\sec u|+ C[\/latex]<\/li>\n
              • [latex]\\int \\sec u du = \\ln |\\sec u + \\tan u|+ C[\/latex]<\/li>\n
              • [latex]\\int \\frac{1}{a^2+u^2} du = \\frac{1}{a}\\arctan \\left( \\frac{u}{a} \\right)+ C[\/latex]<\/li>\n
              • [latex]\\int \\frac{1}{\\sqrt{a^2-u^2}} du = \\arcsin \\left( \\frac{u}{a} \\right)+ C[\/latex]<\/li>\n<\/ul>\n<\/section>\n
                Calculate each of the following integrals:<\/p>\n
                  \n
                1. [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
                2. [latex]\\displaystyle\\int_{} \\big[{\\langle}t,\\ t^{2},\\ t^{3}\\rangle\\times{\\langle}t^{3},\\ t^{2},\\ t\\rangle\\big]\\,dt[\/latex]<\/li>\n
                3. [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
                  \n