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]
TRY IT
Calculate the following integral:
[latex]\displaystyle\int_{1}^{3} \ \big[(2t+4)\,{\bf{i}}+(3t^{2}-4t)\,{\bf{j}}\big]\,dt[/latex]
Watch the following video to see the worked solution to the above Try It
Candela Citations
- CP 3.8. Authored by: Ryan Melton. License: CC BY: Attribution
- Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/books/calculus-volume-3/pages/1-introduction. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction