Integrals of Vector-Valued Functions

# Integrals of Vector-Valued Functions

We have just looked at computing Derivatives of Vector-Valued Functions. We will now look at the vector-valued function analogue of integration. Let's first start off with indefinite integration.

## Indefinite Integration of Vector-Valued Functions

 Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function. Then $\vec{R}(t)$ is said to be an Antiderivative of $\vec{r}(t)$ if $\vec{R'}(t) = \vec{r}(t)$.

As we see, the notion of an antiderivative of a vector-valued function is analogous to that of an antiderivative of a real-valued function.

 Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function defined for all $t$ in the interval $I$. If $x(t)$, $y(t)$, and $z(t)$ are integrable on $I$, then $\int \vec{r}(t) \: dt = \left ( \int x(t) \: dt, \int y(t) \: dt, \int z(t) \: dt \right ) + \vec{C}$. We call $\vec{C}$ the Vector Constant of Integration.

One important note to mention is that the constant of integration $\vec{C}$ is a vector when integrating vector-valued functions. In fact, $\vec{C} = (C_1, C_2, C_3)$ is the vector whose components are the constants of integration when integrating $x(t)$, $y(t)$, and $z(t)$ respectively.

Now that that's aside, let's look at an example of integrating a vector-valued function. Consider the vector-valued function $\vec{r}(t) = (2 \cos t, t^3, e^t - 1)$. To integrate $\vec{r}(t)$, all we need to do is integrate each component function as follows:

(1)
\begin{align} \quad \int \vec{r}(t) \: dt = \left ( \int 2 \cos t \: dt, \int t^3 \: dt, \int e^t - 1 \: dt \right ) + \vec{C} = \left ( 2 \sin t, \frac{t^4}{4}, e^t - t \right ) + \vec{C} \end{align}

Of course, if the components of $\vec{r}(t)$ are more complicated to integrate, we can always utilize techniques we have learned for real-valued functions for integration, such as U-Substitution, Integration by Parts, Tabular Integration, etc…

Now let's look at definite integration of vector-valued functions.

## Definite Integration of Vector-Valued Functions

 Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function defined for all $t$ in the interval $I$. If $x(t)$, $y(t)$, and $z(t)$ are integrable on $I$, then $\int_a^b \vec{r}(t) \: dt = \left ( \int_a^b x(t) \: dt, \int_a^b y(t) \: dt , \int_a^b z(t) \: dt \right )$.

An alternative way to denote definite integration of a vector-valued function is $\int_a^b \vec{r}(t) \: dt = \left ( \int x(t) \: dt, \int y(t) \: dt, \int z(t) \: dt \right ) \biggr \rvert _a^b$, which is often neater to write.

 Theorem 1: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function defined for all $t$ in the interval $I$. If $x(t)$, $y(t)$, and $z(t)$ are integrable on $I$, then $\int_a^b \vec{r}(t) = \vec{R}(b) - \vec{R}(a)$ where $\vec{R}(t)$ is any antiderivative to $\vec{r}(t)$.
• Proof: From the definition of integrating vector-valued functions we have that:
(2)
\begin{align} \quad \int_a^b \vec{r}(t) \: dt = \left ( \int_a^b x(t) \: dt, \int_a^b y(t) \: dt, int_a^b z(t) \:dt \right ) \\ \quad \int_a^b \vec{r}(t) \: dt = \left ( [X(t) + C_1] \biggr \rvert _a^b , [Y(t) + C_2] \biggr \rvert _a^b, [Z(t) + C_3] \biggr \rvert _a^b \right ) \\ \quad \int_a^b \vec{r}(t) \: dt = \left ( X(b) - X(a), Y(b) - Y(a), Z(b) - Z(a) \right ) \\ \quad \int_a^b \vec{r}(t) \: dt = (X(b), Y(b), Z(b)) - (X(a), Y(a), Z(a)) \\ \quad \int_a^b \vec{r}(t) \: dt = \vec{R}(b) - \vec{R}(a) \quad \blacksquare \end{align}

Now let's look at an example of definite integration of a vector-valued function.

Consider the vector-valued function $\vec{r}(t) = ( \sin t, t, 3e^t )$ that we want to integrate from $0$ to $2$. If we indefinitely integrate this function, we get that $\int \vec{r}(t) \: dt = \left ( \int \sin t \: dt, \int t \: dt , \int 3e^t \: dt \right ) = (-\cos t, \frac{t^2}{2}, 3e^t)$ and so it follows that:

(3)
\begin{align} \quad \int_0^2 \vec{r}(t) \: dt = (-\cos t, \frac{t^2}{2}, 3e^t) \biggr \rvert_0^2 \\ \quad \int_0^2 \vec{r}(t) \: dt = (-\cos 2, 2, 3e^2) - (-1, 0, 3) = (1 - \cos 2, 2, 3(e^2 + 1)) \end{align}
 Note: Recall that if we have a real-valued integrable function $f$, then the indefinite integral of $f$ produces a function that is an antiderivative of $f$, while a definite integral of $f$ produces a number. When dealing with vector-valued integrable functions, the indefinite integral of $\vec{r}(t)$ produces a vector function $\vec{R}(t)$, while a definite integral of $\vec{r}(t)$ produces a single vector as you should notice from the last two examples.