Definition of Vector-Valued Functions
A function of the form
\[\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j} + h\left( t \right)\mathbf{k}\;\;\;\text{or}\;\;\;\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \]
is called a vector-valued function in \(3D\) space, where \(f\left( t \right),\) \(g\left( t \right),\) \(h\left( t \right)\) are the component functions depending on the parameter \(t.\)
We can likewise define a vector-valued function in \(2D\) space (in plane):
\[\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j}\;\;\;\text{or}\;\;\;\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right)} \right\rangle .\]
Antiderivatives of Vector-Valued Functions
The vector-valued function \(\mathbf{R}\left( t \right)\) is called an antiderivative of the vector-valued function \(\mathbf{r}\left( t \right)\) whenever
\[\mathbf{R}^\prime\left( t \right) = \mathbf{r}\left( t \right).\]
In component form, if \(\mathbf{R}\left( t \right) = \left\langle {F\left( t \right),G\left( t \right),H\left( t \right)} \right\rangle \) and \(\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle,\) then
\[\left\langle {F^\prime\left( t \right),G^\prime\left( t \right),H^\prime\left( t \right)} \right\rangle = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle .\]
Note that the vector function
\[\left\langle {F\left( t \right) + {C_1},\,G\left( t \right) + {C_2},\,H\left( t \right) + {C_3}} \right\rangle \]
is also an antiderivative of \(\mathbf{r}\left( t \right)\).
The \(3\) scalar constants \({C_1},{C_2},{C_3}\) produce one vector constant, so the most general antiderivative of \(\mathbf{r}\left( t \right)\) has the form
\[{\mathbf{R}\left( t \right)} + \mathbf{C},\]
where \(\mathbf{C} = \left\langle {{C_1},{C_2},{C_3}} \right\rangle .\)
Indefinite Integral of a Vector-Valued Function
If \(\mathbf{R}\left( t \right)\) is an antiderivative of \(\mathbf{r}\left( t \right),\) the indefinite integral of \(\mathbf{r}\left( t \right)\) is
\[\int {\mathbf{r}\left( t \right)dt} = \mathbf{R}\left( t \right) + \mathbf{C},\]
where \(\mathbf{C}\) is an arbitrary constant vector.
In component form, the indefinite integral is given by
\[\int {\mathbf{r}\left( t \right)dt} = \int {\left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle dt} = \left\langle {\int {f\left( t \right)dt} ,\int {g\left( t \right)dt} ,\int {h\left( t \right)dt} } \right\rangle.\]
Definite Integral of a Vector-Valued Function
The definite integral of \(\mathbf{r}\left( t \right)\) on the interval \(\left[ {a,b} \right]\) is defined by
\[\int\limits_a^b {\mathbf{r}\left( t \right)dt} = \int\limits_a^b {\left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle dt} = \left\langle {\int\limits_a^b {f\left( t \right)dt} ,\int\limits_a^b {g\left( t \right)dt} ,\int\limits_a^b {h\left( t \right)dt} } \right\rangle.\]
We can extend the Fundamental Theorem of Calculus to vector-valued functions.
If \(\mathbf{r}\left( t \right)\) is continuous on \(\left( {a,b} \right),\) then
\[\int\limits_a^b {\mathbf{r}\left( t \right)dt} = \mathbf{R}\left( b \right) - \mathbf{R}\left( a \right),\]
where \(\mathbf{R}\left( t \right)\) is any antiderivative of \(\mathbf{r}\left( t \right).\)
Vector-valued integrals obey the same linearity rules as scalar-valued integrals.
Solved Problems
Click or tap a problem to see the solution.
Example 1
Evaluate the integral \[\int\limits_0^{\frac{\pi }{2}} {\left\langle {\sin t,2\cos t,1} \right\rangle dt}.\]
Example 2
Find the integral \[\int {\left( {{{\sec }^2}t\mathbf{i} + \ln t\mathbf{j}} \right)dt}.\]
Example 3
Find the integral \[\int {\left( {\frac{1}{{{t^2}}} \mathbf{i} + \frac{1}{{{t^3}}} \mathbf{j} + t\mathbf{k}} \right)dt}.\]
Example 4
Evaluate the indefinite integral \[\int {\left\langle {4\cos 2t,4t{e^{{t^2}}},2t + 3{t^2}} \right\rangle dt}.\]
Example 5
Evaluate the indefinite integral \[\int {\left\langle {\frac{1}{t},4{t^3},\sqrt t } \right\rangle dt},\] where \(t \gt 0.\)
Example 6
Find \(\mathbf{R}\left( t \right)\) if \[\mathbf{R}^\prime\left( t \right) = \left\langle {1 + 2t,2{e^{2t}}} \right\rangle \] and \(\mathbf{R}\left( 0 \right) = \left\langle {1,3} \right\rangle .\)
Example 1.
Evaluate the integral \[\int\limits_0^{\frac{\pi }{2}} {\left\langle {\sin t,2\cos t,1} \right\rangle dt}.\]
Solution.
By integrating componentwise, we have
\[\int\limits_0^{\frac{\pi }{2}} {\left\langle {\sin t,2\cos t,1} \right\rangle dt} = \left\langle {{\int\limits_0^{\frac{\pi }{2}} {\sin tdt}} ,{\int\limits_0^{\frac{\pi }{2}} {2\cos tdt}} ,{\int\limits_0^{\frac{\pi }{2}} {1dt}} } \right\rangle = \left\langle {\left. { - \cos t} \right|_0^{\frac{\pi }{2}},\left. {2\sin t} \right|_0^{\frac{\pi }{2}},\left. t \right|_0^{\frac{\pi }{2}}} \right\rangle = \left\langle {0 + 1,2 - 0,\frac{\pi }{2} - 0} \right\rangle = \left\langle {{1},{2},{\frac{\pi }{2}}} \right\rangle .\]
Example 2.
Find the integral \[\int {\left( {{{\sec }^2}t\mathbf{i} + \ln t\mathbf{j}} \right)dt}.\]
Solution.
We integrate on a component-by-component basis:
\[I = \int {\left( {{{\sec }^2}t\mathbf{i} + \ln t\mathbf{j}} \right)dt} = \left( {\int {{{\sec }^2}tdt} } \right)\mathbf{i} + \left( {\int {\ln td} t} \right)\mathbf{j}.\]
The first integral is given by
\[\int {{{\sec }^2}tdt} = \tan t.\]
The second integral can be computed using integration by parts:
\[\int {\ln td} t = \left[ {\begin{array}{*{20}{l}}
{u = \ln t}\\
{dv = dt}\\
{du = \frac{1}{t}dt}\\
{v = t}
\end{array}} \right] = t\ln t - \int {t \cdot \frac{1}{t}dt} = t\ln t - \int {dt} = t\ln t - t = t\left( {\ln t - 1} \right).\]
Thus, the given integral is equal to
\[I = \tan t\mathbf{i} + t\left( {\ln t - 1} \right)\mathbf{j} + \mathbf{C},\]
where \(\mathbf{C} = {C_1}\mathbf{i} + {C_2}\mathbf{j}\) is an arbitrary constant vector.
Example 3.
Find the integral \[\int {\left( {\frac{1}{{{t^2}}} \mathbf{i} + \frac{1}{{{t^3}}} \mathbf{j} + t\mathbf{k}} \right)dt}.\]
Solution.
Integrating on a component-by-component basis yields:
\[\int {\left( {\frac{1}{{{t^2}}}\mathbf{i} + \frac{1}{{{t^3}}}\mathbf{j} + t\mathbf{k}} \right)dt} = \left( {\int {\frac{{dt}}{{{t^2}}}} } \right)\mathbf{i} + \left( {\int {\frac{{dt}}{{{t^3}}}} } \right)\mathbf{j} + \left( {\int {tdt} } \right)\mathbf{k} = \left( {\int {{t^{ - 2}}dt} } \right)\mathbf{i} + \left( {\int {{t^{ - 3}}dt} } \right)\mathbf{j} + \left( {\int {tdt} } \right)\mathbf{k} = \frac{{{t^{ - 1}}}}{{\left( { - 1} \right)}}\mathbf{i} + \frac{{{t^{ - 2}}}}{{\left( { - 2} \right)}}\mathbf{j} + \frac{{{t^2}}}{2}\mathbf{k} + \mathbf{C} = - \frac{1}{t}\mathbf{i} - \frac{1}{{2{t^2}}}\mathbf{j} + \frac{{{t^2}}}{2}\mathbf{k} + \mathbf{C},\]
where \(\mathbf{C} = {C_1}\mathbf{i} + {C_2}\mathbf{j}\) is a constant vector.
Example 4.
Evaluate the indefinite integral \[\int {\left\langle {4\cos 2t,4t{e^{{t^2}}},2t + 3{t^2}} \right\rangle dt}.\]
Solution.
Integrating componentwise yields:
\[I = \int {\left\langle {4\cos 2t,4t{e^{{t^2}}},2t + 3{t^2}} \right\rangle dt} = \left\langle {\int {4\cos 2tdt} ,\int {4t{e^{{t^2}}}dt} ,\int {\left( {2t + 3{t^2}} \right)dt} } \right\rangle .\]
We evaluate each integral separately.
The first integral is given by
\[\int {4\cos 2tdt} = 4 \cdot \frac{{\sin 2t}}{2} + {C_1} = 2\sin 2t + {C_1}.\]
To compute the second integral, we make the substitution \(u = {t^2},\) \(du = 2tdt.\) Then
\[\int {4t{e^{{t^2}}}dt} = 2\int {{e^u}du} = 2{e^u} + {C_2} = 2{e^{{t^2}}} + {C_2}.\]
The third integral is pretty straightforward:
\[\int {\left( {2t + 3{t^2}} \right)dt} = {t^2} + {t^3} + {C_3}.\]
Thus, the initial integral is equal
\[I = \left\langle {2\sin 2t + {C_1},\,2{e^{{t^2}}} + {C_2},\,{t^2} + {t^3} + {C_3}} \right\rangle = \left\langle {2\sin 2t,2{e^{{t^2}}},{t^2} + {t^3}} \right\rangle + \left\langle {{C_1},{C_2},{C_3}} \right\rangle = \left\langle {2\sin 2t,2{e^{{t^2}}},{t^2} + {t^3}} \right\rangle + \mathbf{C},\]
where \(\mathbf{C} = \left\langle {{C_1},{C_2},{C_3}} \right\rangle \) is an arbitrary constant vector.
Example 5.
Evaluate the indefinite integral \[\int {\left\langle {\frac{1}{t},4{t^3},\sqrt t } \right\rangle dt},\] where \(t \gt 0.\)
Solution.
We integrate component-by-component:
\[\int {\left\langle {\frac{1}{t},4{t^3},\sqrt t } \right\rangle dt} = \left\langle {\int {\frac{{dt}}{t}} ,\int {4{t^3}dt} ,\int {\sqrt t dt} } \right\rangle = \left\langle {\ln t,{t^4},\frac{{2\sqrt {{t^3}} }}{3}} \right\rangle + \left\langle {{C_1},{C_2},{C_3}} \right\rangle = \left\langle {\ln t,3{t^4},\frac{{3\sqrt {{t^3}} }}{2}} \right\rangle + \mathbf{C},\]
where \(\mathbf{C} = \left\langle {{C_1},{C_2},{C_3}} \right\rangle \) is any number vector.
Example 6.
Find \(\mathbf{R}\left( t \right)\) if \[\mathbf{R}^\prime\left( t \right) = \left\langle {1 + 2t,2{e^{2t}}} \right\rangle \] and \(\mathbf{R}\left( 0 \right) = \left\langle {1,3} \right\rangle .\)
Solution.
First we integrate the vector-valued function:
\[\mathbf{R}\left( t \right) = \int {\left\langle {1 + 2t,2{e^{2t}}} \right\rangle dt} = \left\langle {\int {\left( {1 + 2t} \right)dt} ,\int {2{e^{2t}}dt} } \right\rangle = \left\langle {t + {t^2},{e^{2t}}} \right\rangle + \left\langle {{C_1},{C_2}} \right\rangle = \left\langle {t + {t^2},{e^{2t}}} \right\rangle + \mathbf{C}.\]
We determine the vector \(\mathbf{C}\) from the initial condition \(\mathbf{R}\left( 0 \right) = \left\langle {1,3} \right\rangle :\)
\[\mathbf{R}\left( 0 \right) = \left\langle {0 + {0^2},{e^0}} \right\rangle + \mathbf{C} = \left\langle {0,1} \right\rangle + \mathbf{C} = \left\langle {1,3} \right\rangle .\]
Hence
\[\mathbf{C} = \left\langle {1,3} \right\rangle - \left\langle {0,1} \right\rangle = \left\langle {1,2} \right\rangle .\]
The answer is given by
\[\mathbf{R}\left( t \right) = \left\langle {t + {t^2},{e^{2t}}} \right\rangle + \left\langle {1,2} \right\rangle .\]
