Integrals of Vector-Valued Functions

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 (t), g (t), h (t) 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

$\mathbf{C} = \left\langle {1,3} \right\rangle - \left\langle {0,1} \right\rangle = \left\langle {1,2} \right\rangle .$

$\mathbf{R}\left( t \right) = \left\langle {t + {t^2},{e^{2t}}} \right\rangle + \left\langle {1,2} \right\rangle .$