Integrals of Vector-Valued Functions

Learning Outcomes

Calculate the definite integral of a vector-valued function

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.

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 t, 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.

Definition

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

The indefinite integral of a vector-valued function [latex]{\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}[/latex] is
[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].

The definite integral of a vector-valued function is

[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]
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
[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]

The definite integral of the vector-valued function is

[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]

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

[latex]\displaystyle\int_{} \ f\,(t)\,dt=F\,(t)+C_{1}[/latex] and [latex]\displaystyle\int_{} \ g\,(t)\,dt=G\,(t)+C_{2}[/latex],

where F and G are antiderivatives of [latex]f[/latex] and [latex]g[/latex], respectively. Then

[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]

where [latex]C=C_{1}\,{\bf{i}}+C_{2}\,{\bf{j}}[/latex]. Therefore, the integration constant becomes a constant vector.

Since we will also encounter integrals frequently throughout this course, we review several common integrals below.

Recall: Integrals of common functions

- [latex] \int x^n \ dx = \frac{1}{n+1}x^{n+1} + C \ (\text{ if } x \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 u \tan u \ du = \sec u+ C [/latex]
- [latex] \int \csc u \cot u \ du = -\csc 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 \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]

Example: integrating vector-valued functions

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]

Show Solution

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]

TRY IT

Calculate the following integral:

[latex]\displaystyle\int_{1}^{3} \ \big[(2t+4)\,{\bf{i}}+(3t^{2}-4t)\,{\bf{j}}\big]\,dt[/latex]

Show Solution

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 3.8” here (opens in new window).

Module 3: Vector-Valued Functions