Question

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=\left\langle t^{3}-3 t, 2 t-1,10\right\rangle$$

Short answer

Answer: The indefinite integral of the vector function \( \mathbf{r}(t) \) is given by \( \mathbf{R}(t) = \left\langle \frac{1}{4}t^4 - \frac{3}{2}t^2 + C_1, \ t^2 - t + C_2, \ 10t + C_3 \right\rangle \), where \( C_1 \), \( C_2 \) and \( C_3 \) are constants of integration.

Step-by-step solution

Identify Components

The provided vector function has three components: 1. \(x(t) = t^3 - 3t\) 2. \(y(t) = 2t - 1\) 3. \(z(t) = 10\)

Integrate Each Component

Integrate each component with respect to \(t\): 1. \(\int (t^3-3t) dt\) 2. \(\int (2t-1) dt\) 3. \(\int 10 dt\)

Solve Component Integrals

After solving each integral, we get: 1. \(\int (t^3-3t) dt = \frac{1}{4}t^4 - \frac{3}{2}t^2 + C_1\) 2. \(\int (2t-1) dt = t^2 - t + C_2\) 3. \(\int 10 dt = 10t + C_3\)

Write the Indefinite Integral

Combine these component integrals to form the final indefinite integral: $$ \mathbf{R}(t) = \left\langle \frac{1}{4}t^4 - \frac{3}{2}t^2 + C_1, \ t^2 - t + C_2, \ 10t + C_3 \right\rangle $$

Key concepts

Vector Calculus

Vector calculus is a field of mathematics that deals with vector fields and the differentiation and integration of vector functions. In this exercise, we dealt with a vector function \( \mathbf{r}(t) = \langle t^3 - 3t, 2t - 1, 10 \rangle \). Vector functions like \( \mathbf{r}(t) \) are composed of multiple component functions—each representing a different dimension in space. Vector calculus allows us to analyze these functions and understand their behavior in multi-dimensional space.

Common operations in vector calculus include differentiation and integration of vector functions, just like we do with scalar functions. Here, our task was to compute the indefinite integrals of each component of \( \mathbf{r}(t) \), bringing us to the realm of vector integration. This enables us to find a function that represents the accumulation of change along a vector field.

Component Integration

Component integration is a technique used when working with vector functions. Since vector functions are composed of multiple individual functions, each corresponding to a component of the vector, they must be integrated separately.

In our exercise, the vector function \( \mathbf{r}(t) = \langle t^3 - 3t, 2t - 1, 10 \rangle \) was broken down into three components:

• \( x(t) = t^3 - 3t \)
• \( y(t) = 2t - 1 \)
• \( z(t) = 10 \)

Each component is integrated separately with respect to \( t \). This method ensures each part of the vector is accurately processed:

• \( \int (t^3 - 3t) \ dt = \frac{1}{4}t^4 - \frac{3}{2}t^2 + C_1 \)
• \( \int (2t - 1) \ dt = t^2 - t + C_2 \)
• \( \int 10 \ dt = 10t + C_3 \)

Remember to add constants of integration \( C_1, C_2, C_3 \) since these integrals are indefinite.

Integration Technique

The integration technique used in this exercise is straightforward but crucial for solving indefinite integrals for vector fields. In simple terms, integration is the process of finding a function that describes the accumulation of quantities. Here, we focus on the indefinite integral, which results in a function plus a constant (or constants, in the vector case).

To integrate functions like \( \int (t^3 - 3t) \ dt \) or \( \int (2t - 1) \ dt \), we apply basic integration rules such as the power rule and linearity of integration:

• The power rule states that \( \int t^n \ dt = \frac{1}{n+1}t^{n+1} + C \), which is applied when integrating terms like \( t^3 \).
• For linear terms such as \( 2t \), you simply adjust constants, giving \( \int 2t \ dt = t^2 + C \).
• Constant terms are integrated by multiplying by \( t \), as in \( \int 10 \ dt = 10t + C \).

By integrating each component separately using these basic techniques, you can reconstruct the vector's indefinite integral, maintaining the initial vector's structure but accounting for the cumulative effect over a variable \( t \).