Calculus of Vector-Valued Functions: Learn It 3

Giselle Miller

Calculus of Vector-Valued Functions: Learn It 3

Integrals of Vector-Valued Functions

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.

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.

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.

integrals of vector-valued functions

Let [latex]f[/latex], [latex]g[/latex], and [latex]h[/latex] be integrable real-valued functions over the closed interval [latex][a, b][/latex].

For plane curves:

Indefinite integral: [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]
Definite integral: [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]

For space curves:

Indefinite integral: [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]
Definite integral: [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]

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].

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:

[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]

where [latex]{\bf{C}}=C_{1}{\bf{i}}+C_{2}{\bf{j}}[/latex].

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.

Integrals of Common Functions

[latex]\int x^n dx = \frac{1}{n+1}x^{n+1} + C[/latex] (if [latex]n \ne -1[/latex])
[latex]\int \frac{1}{x} dx = \ln |x|+ C[/latex]
[latex]\int e^u du = e^u+ C[/latex]
[latex]\int \cos u du = \sin u+ C[/latex]
[latex]\int \sin u du = -\cos u+ C[/latex]
[latex]\int \sec^2 u du = \tan u+ C[/latex]
[latex]\int \csc^2 u du = -\cot u+ C[/latex]
[latex]\int \sec u \tan u du = \sec u+ C[/latex]
[latex]\int \csc u \cot u du = -\csc u+ C[/latex]
[latex]\int \tan u du = \ln |\sec u|+ C[/latex]
[latex]\int \sec u du = \ln |\sec u + \tan u|+ C[/latex]
[latex]\int \frac{1}{a^2+u^2} du = \frac{1}{a}\arctan \left( \frac{u}{a} \right)+ C[/latex]
[latex]\int \frac{1}{\sqrt{a^2-u^2}} du = \arcsin \left( \frac{u}{a} \right)+ C[/latex]

Calculate each of the following integrals:

[latex]\displaystyle\int_{} \big[(3t^{2}+2t)\,{\bf{i}}+(3t-6)\,{\bf{j}}+(6t^{3}+5t^{2}-4)\,{\bf{k}}\big]\,dt[/latex]
[latex]\displaystyle\int_{} \big[{\langle}t,\ t^{2},\ t^{3}\rangle\times{\langle}t^{3},\ t^{2},\ t\rangle\big]\,dt[/latex]
[latex]\displaystyle\int_{0}^{\frac{\pi}{3}}\big[\sin{2t\,{\bf{i}}}+\tan{t\,{\bf{j}}}+e^{-2t}\,{\bf{k}}\big]\,dt[/latex]

a. We use the first part of the definition of the integral of a space curve:

[latex]\displaystyle\int_{} \ \big[(3t^{2}+2t)\,{\bf{i}}+(3t-6)\,{\bf{j}}+(6t^{3}+5t^{2}-4)\,{\bf{k}}\big]\,dt[/latex]

[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]

b. First calculate [latex]\langle{t},\ t^{2},\ t^{3}\rangle\times\langle{t}^{3},\ t^{2},\ {t}\rangle[/latex]:

[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]

Next, substitute this back into the integral and integrate:

[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]

c. Use the second part of the definition of the integral of a space curve:

[latex]\displaystyle\int_{0}^{\frac{\pi}{3}} \big[\sin{2t\,{\bf{i}}}+\tan{t\,{\bf{j}}}+e^{-2t}\,{\bf{k}}\big]\,dt[/latex]

[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]