Question

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=t e^{t} \mathbf{i}+t \sin t^{2} \mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}} \mathbf{k}$$

Short answer

Question: Find the indefinite integral of the following vector-valued function: $$\mathbf{r}(t)=t e^{t} \mathbf{i}+t \sin t^{2} \mathbf{j}-\frac{2t}{\sqrt{t^{2}+4}} \mathbf{k}$$ Answer: The indefinite integral of the given vector-valued function is: $$\int \textbf{r}(t) dt = (te^t - e^t + C_1)\textbf{i} -\frac{1}{2}\cos (t^2) + C_2 \textbf{j} -2\sqrt{(t^2+4)} + C_3 \textbf{k}$$

Step-by-step solution

Identify the functions to integrate

We have a vector-valued function as follows: $$\mathbf{r}(t)=t e^{t} \mathbf{i}+t \sin t^{2} \mathbf{j}-\frac{2t}{\sqrt{t^{2}+4}} \mathbf{k}$$ To compute the indefinite integral of \(\textbf{r}(t)\), we need to find the indefinite integral of each of its components: \(t e^{t}\), \(t \sin t^{2}\), and \(-\frac{2t}{\sqrt{t^{2}+4}}\).

Integrate the \(\textbf{i}\) component

We need to integrate the following function: $$t e^{t}$$ To perform this integration, we will use integration by parts. Let \(u=t\) and \(dv=e^{t}dt\). Then \(du = dt\) and \(v = \int e^t dt = e^t\). Now apply integration by parts formula: $$\int t e^{t} dt = uv - \int v du = te^t - \int e^t dt = te^t - e^t + C_1$$ where \(C_1\) is the constant of integration.

Integrate the \(\textbf{j}\) component

We need to integrate the following function: $$t \sin t^{2}$$ We will use substitution method to integrate this function. Let \(u=t^2\), then \(du=2tdt\). Thus, $$\int t \sin t^{2} dt = \frac{1}{2}\int \sin u du = -\frac{1}{2}\cos u + C_2=-\frac{1}{2}\cos t^2 + C_2$$ where \(C_2\) is the constant of integration.

Integrate the \(\textbf{k}\) component

We need to integrate the following function: $$-\frac{2t}{\sqrt{t^{2}+4}}$$ We will use substitution method again. Let \(u=t^2 + 4\), then \(du=2t dt\). So, $$\int -\frac{2t}{\sqrt{t^{2}+4}} dt = -\int \frac{du}{\sqrt{u}} = -\int u^{-\frac{1}{2}} du = -2\sqrt{u} + C_3 = -2\sqrt{t^2+4} + C_3$$ where \(C_3\) is the constant of integration.

Combine the indefinite integrals for final result

Now that we have the indefinite integral of each individual component, we can combine them to obtain the indefinite integral of the vector-valued function \(\textbf{r}(t)\): $$\int \textbf{r}(t) dt = (te^t - e^t + C_1)\textbf{i} -\frac{1}{2}\cos t^2 + C_2 \textbf{j} -2\sqrt{t^2+4} + C_3 \textbf{k}$$ So, the indefinite integral of the given vector-valued function is: $$\int \textbf{r}(t) dt = (te^t - e^t + C_1)\textbf{i} -\frac{1}{2}\cos (t^2) + C_2 \textbf{j} -2\sqrt{(t^2+4)} + C_3 \textbf{k}$$

Key concepts

Indefinite Integral

An indefinite integral, also known as an antiderivative, is a fundamental concept in calculus. It represents a family of functions whose derivative gives the original function. The indefinite integral is expressed as \( \int f(x) \, dx \), where \( f(x) \) is the function you want to integrate. The result includes a constant of integration, typically denoted as \( C \), because adding any constant value would still satisfy the derivative conditions.

This concept is a crucial tool for recovering a function from its rate of change. In our exercise, we're computing the indefinite integral of a vector-valued function. This involves individually integrating each component function of the vector, such as \( t e^{t} \), \( t \sin t^{2} \), and \( -\frac{2t}{\sqrt{t^{2}+4}} \). Each component has its own indefinite integral and constant of integration.
Understanding how to compute indefinite integrals is essential in many fields, including physics and engineering. They help describe systems involving motion, growth, or change over time.

Integration By Parts

Integration by parts is a technique used for finding integrals of products of functions. It is based on the product rule for differentiation and has a specific formula: \( \int u \, dv = uv - \int v \, du \). Here, \( u \) and \( dv \) are parts that you split the original function into.

In our vector calculus problem, we utilize integration by parts to solve the integral of the \( \textbf{i} \) component: \( \int t e^{t} \, dt \). For this, we let \( u = t \) and \( dv = e^{t} \, dt \). This choice simplifies the differentiation of \( t \) and the integration of \( e^{t} \).
Then, we compute \( du = dt \) and \( v = e^{t} \). Applying the integration by parts formula results in the computation: \( te^{t} - e^{t} + C_1 \). This method is powerful for functions where traditional integration methods fail, particularly when dealing with polynomial times exponential forms.

Substitution Method

The substitution method is another critical technique for integrating functions that appear complex at first glance. It involves substituting part of the integrand (the function being integrated) with a single variable, typically denoted as \( u \), to simplify the integral.

This method has been applied to both the \( \textbf{j} \) and \( \textbf{k} \) components in the vector-valued function: \( \int t \sin t^{2} \, dt \) and \( \int -\frac{2t}{\sqrt{t^{2}+4}} \, dt \). For \( \int t \sin t^{2} \, dt \), \( u = t^2 \) is set, simplifying the integral with \( du = 2t \, dt \). The integration becomes \( -\frac{1}{2}\cos t^{2} + C_2 \).
For \( \int -\frac{2t}{\sqrt{t^{2}+4}} \, dt \), \( u = t^2 + 4 \) was chosen, giving \( du = 2t \, dt \). This leads to \( -2\sqrt{t^{2}+4} + C_3 \). Substitution makes it easier to integrate functions involving compositions of more complex expressions.

Vector-Valued Function

A vector-valued function is a function that outputs a vector, rather than a single scalar value. In other words, for an input \( t \), it returns a vector often composed of components along the \( \textbf{i} \), \( \textbf{j} \), and \( \textbf{k} \) axes.

In our specific exercise, \( \textbf{r}(t) = t e^{t} \textbf{i} + t \sin t^{2} \textbf{j} - \frac{2t}{\sqrt{t^{2}+4}} \textbf{k} \) represents such a function. Each component describes a behavior constrained to its directional axis, making this type of function particularly important in physics for describing paths in three-dimensional space.
To find the indefinite integral of a vector-valued function, you integrate each component individually, as shown in this problem. This approach simplifies complex vector calculus exercises into manageable parts, allowing you to understand each direction's contribution separately before combining them into a final vector result.