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

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

The definite integral of a vector-valued function is

$\displaystyle\int_{a}^{b} \ [f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}]\,dt=\Bigg[\displaystyle\int_{a}^{b} \ f\,(t)\,dt\Bigg]{\bf{i}}+\Bigg[\displaystyle\int_{a}^{b} \ g\,(t)\,dt\Bigg]{\bf{j}}$
The indefinite integral of a vector-valued function ${\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}$ is
$\displaystyle\int_{} \ [f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}]\,dt=\Bigg[\displaystyle\int_{} \ f\,(t)\,dt\Bigg]{\bf{i}}+\Bigg[\displaystyle\int_{} \ g\,(t)\,dt\Bigg]{\bf{j}}+\Bigg[\displaystyle\int_{} \ h\,(t)\,dt\Bigg]\,{\bf{k}}$

The definite integral of the vector-valued function is

$\displaystyle\int_{a}^{b} \ [f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}]\,dt=\Bigg[\displaystyle\int_{a}^{b} \ f\,(t)\,dt\Bigg]{\bf{i}}+\Bigg[\displaystyle\int_{a}^{b} \ g\,(t)\,dt\Bigg]{\bf{j}}+\Bigg[\displaystyle\int_{a}^{b} \ h\,(t)\,dt\Bigg]\,{\bf{k}}.$

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

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

where F and G are antiderivatives of $f$ and $g$ , respectively. Then

$\begin{array}{ccc}\hfill {\displaystyle\int_{} \ [f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}]\,dt} & =\hfill & {\Bigg[\displaystyle\int_{} \ f\,(t)\,dt\Bigg]\,{\bf{i}}+\Bigg[\int_{} \ g\,(t)\,dt\Bigg]\,{\bf{j}}} \hfill \\ \hfill & =\hfill & {(F\,(t)+C_{1})\,{\bf{i}}+(G\,(t)+C_{2})\,{\bf{j}}} \hfill \\ \hfill & =\hfill & {F\,(t)\,{\bf{i}}+G\,(t)\,{\bf{j}}+C_{1}\,{\bf{i}}+C_{2}\,{\bf{j}}} \hfill \\ \hfill & =\hfill & {F\,(t)\,{\bf{i}}+G\,(t)\,{\bf{j}}+C,} \hfill \\ \hfill \end{array}$

where $C=C_{1}\,{\bf{i}}+C_{2}\,{\bf{j}}$ . 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

- $\int x^n \ dx = \frac{1}{n+1}x^{n+1} + C \ (\text{ if } x \ne -1)$
- $\int \frac{1}{x} \ dx = \ln |x|+ C$
- $\int e^u \ du = e^u+ C$
- $\int \cos u \ du = \sin u+ C$
- $\int \sin u \ du = - \cos u+ C$
- $\int \sec u \tan u \ du = \sec u+ C$
- $\int \csc u \cot u \ du = -\csc u+ C$
- $\int \sec^2 u \ du = \tan u+ C$
- $\int \csc^2 u \ du = -\cot u+ C$

- $\int \tan u \ du = \ln |\sec u|+ C$
- $\int \sec u \ du = \ln |\sec u + \tan u]+ C$
- $\int \frac{1}{a^2+u^2} \ du = \frac{1}{a}\arctan \left( \frac{u}{a} \right)+ C$
- $\int \frac{1}{\sqrt{a^2-u^2}} \ du = \arcsin \left( \frac{u}{a} \right)+ C$

Example: integrating vector-valued functions

Calculate each of the following integrals:

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

Show Solution

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

\int_{} [(3 t^{2} + 2 t) i + (3 t - 6) j + (6 t^{3} + 5 t^{2} - 4) k] d t

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

\begin{array}{ccc} = & [\int_{} 3 t^{2} + 2 t d t] i + [\int_{} 3 t - 6 d t] j + [\int_{} 6 t^{3} + 5 t^{2} - 4 d t] k \\ = & (t^{3} + t^{2}) i + (\frac{3}{2} t^{2} - 6 t) j + (\frac{3}{2} t^{4} + \frac{5}{3} t^{3} - 4 t) k + C . \end{array}

$\begin{array}{ccc}\hfill & =\hfill & {\Big[\displaystyle\int_{} \ 3t^{2}+2t\,dt\Big]\,{\bf{i}}+\Big[\displaystyle\int_{} \ 3t-6\,dt\Big]\,{\bf{j}}+\Big[\displaystyle\int_{} 6t^{3}+5t^{2}-4\,dt\Big]\,{\bf{k}}} \hfill \\ \hfill & =\hfill & {(t^{3}+t^{2})\,{\bf{i}}+(\frac{3}{2}t^{2}-6t)\,{\bf{j}}+(\frac{3}{2}t^{4}+\frac{5}{3}t^{3}-4t)\,{\bf{k}}+C.} \hfill \\ \hfill \end{array}$

b. First calculate $\langle{t},\ t^{2},\ t^{3}\rangle\times\langle{t}^{3},\ t^{2},\ {t}\rangle$ :

\begin{array}{ccc} ⟨ t, t^{2}, t^{3} ⟩ \times ⟨ t^{3}, t^{2}, t ⟩ & = & | \begin{matrix} i & j & k \\ t & t^{2} & t^{3} \\ t^{3} & t^{2} & t \end{matrix} | \\ = & (t^{2} (t) - t^{3} (t^{2})) i - (t^{2} - t^{3} (t^{3})) j + (t (t^{2}) - t^{2} (t^{3})) k \\ = & (t^{3} - t^{5}) i + (t^{6} - t^{2}) j + (t^{3} - t^{5}) k \end{array}

$\begin{array}{ccc}\hfill {\langle{t},\ t^{2},\ t^{3}\rangle\times\langle{t}^{3},\ t^{2},\ t\rangle} & =\hfill & {\begin{vmatrix}{\bf{i}}&{\bf{j}}&{\bf{k}}\\t&t^{2}&t^{3}\\t^{3}&t^{2}&t\end{vmatrix}} \hfill \\ \hfill & =\hfill & {\big(t^{2}\,(t)-t^{3}\,(t^{2})\big)\,{\bf{i}}-\big(t^{2}-t^{3}\,(t^{3})\big)\,{\bf{j}}+\big(t\,(t^{2})-t^{2}\,(t^{3})\big)\,{\bf{k}}} \hfill \\ \hfill & =\hfill & {(t^{3}-t^{5})\,{\bf{i}}+(t^{6}-t^{2})\,{\bf{j}}+(t^{3}-t^{5})\,{\bf{k}}} \hfill \\ \hfill \end{array}$

Next, substitute this back into the integral and integrate:

\begin{array}{ccc} \int_{} [⟨ t, t^{2}, t^{3} ⟩ \times ⟨ t^{3}, t^{2}, t ⟩] d t & = & \int_{} (t^{3} - t^{5}) i + (t^{6} - t^{2}) j + (t^{3} - t^{5}) k d t \\ = & (\frac{t^{4}}{4} - \frac{t^{6}}{6}) i + (\frac{t^{7}}{7} - \frac{t^{3}}{3}) j + (\frac{t^{4}}{4} - \frac{t^{6}}{6}) k + C . \end{array}

$\begin{array}{ccc}\hfill {\displaystyle\int_{} \ \big[\langle{t},\ t^{2},\ t^{3}\rangle\times\langle{t}^{3},\ t^{2},\ t\rangle\big]\,dt} & =\hfill & {\displaystyle\int_{} \ (t^{3}-t^{5})\,{\bf{i}}+(t^{6}-t^{2})\,{\bf{j}}+(t^{3}-t^{5})\,{\bf{k}}\,dt} \hfill \\ \hfill & =\hfill & {\big(\frac{t^{4}}{4}-\frac{t^{6}}{6}\big)\,{\bf{i}}+\big(\frac{t^{7}}{7}-\frac{t^{3}}{3}\big)\,{\bf{j}}+\big(\frac{t^{4}}{4}-\frac{t^{6}}{6}\big)\,{\bf{k}}+C.} \hfill \\ \hfill \end{array}$

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

\int_{0}^{\frac{π}{3}} [\sin 2 t i + \tan t j + e^{- 2 t} k] d t

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

\begin{array}{ccc} = & [\int_{0}^{\frac{π}{3}} \sin 2 t d t] i + [\int_{0}^{\frac{π}{3}} \tan t d t] j + [\int_{0}^{\frac{π}{3}} e^{- 2 t} d t] k \\ = & (- \frac{1}{2} \cos 2 t) |_{0}^{\frac{π}{3}} i - (\ln (\cos t)) |_{0}^{\frac{π}{3}} j - (\frac{1}{2} e^{- 2 t}) |_{0}^{\frac{π}{3}} k \\ = & (- \frac{1}{2} \cos \frac{2 π}{3} + \frac{1}{2} \cos 0)) i - (\ln (\cos \frac{π}{3}) - \ln (\cos 0) j - (\frac{1}{2} e^{- 2 π / 3} - \frac{1}{2} e^{- 2 (0)}) k \\ = & (\frac{1}{4} + \frac{1}{2}) i - (- \ln 2) j - (\frac{1}{2} e^{- 2 π / 3} - \frac{1}{2}) k \\ = & \frac{3}{4} i + (\ln 2) j + (\frac{1}{2} - \frac{1}{2} e^{- 2 π / 3}) k . \end{array}

$\begin{array}{ccc}\hfill & =\hfill & {\Big[\displaystyle\int_{0}^{\frac{\pi}{3}} \sin{2t}\,dt\Big]\,{\bf{i}}+\Big[\displaystyle\int_{0}^{\frac{\pi}{3}} \tan{t}\,dt\Big]\,{\bf{j}}+\Big[\displaystyle\int_{0}^{\frac{\pi}{3}} \ e^{-2t}\,dt\Big]{\bf{k}}} \hfill \\ \hfill & =\hfill & {\big(-\frac{1}{2}\cos{2t}\big)\Big{|}_0^{\frac{\pi}{3}}\,{\bf{i}}-(\ln{(\cos{t})})\Big{|}_0^{\frac{\pi}{3}}\,{\bf{j}}-\big(\frac{1}{2}e^{-2t}\big)\Big{|}_0^{\frac{\pi}{3}}\,{\bf{k}}} \hfill \\ \hfill & =\hfill & {\big(-\frac{1}{2}\cos{\frac{2\pi}{3}}+\frac{1}{2}\cos{0})\big)\,{\bf{i}}-\big(\ln{(\cos{\frac{\pi}{3})}}-\ln{(\cos{0}}\big)\,{\bf{j}}-\big(\frac{1}{2}\,e^{-2\pi/3}-\frac{1}{2}\,e^{-2(0)}\big)\,{\bf{k}}} \hfill \\ \hfill & =\hfill & {\big(\frac{1}{4}+\frac{1}{2}\big)\,{\bf{i}}-(-\ln{2})\,{\bf{j}}-\big(\frac{1}{2}\,e^{-2\pi/3}-\frac{1}{2}\big)\,{\bf{k}}} \hfill \\ \hfill & =\hfill & {\frac{3}{4}\,{\bf{i}}+(\ln{2})\,{\bf{j}}+\big(\frac{1}{2}-\frac{1}{2}\,e^{-2\pi/3}\big)\,{\bf{k}}.} \hfill \\ \hfill \end{array}$

TRY IT

Calculate the following integral:

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

Show Solution

$\displaystyle\int_{1}^{3} \ \big[(2t+4)\,{\bf{i}}+(3t^{2}-4t)\,{\bf{j}}\big]\,dt=16\,{\bf{i}}+10\,{\bf{j}}$

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