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Chapter 12: Problem 54

Find the indefinite integral. $$ \int\left(e^{t} \mathbf{i}+\sin t \mathbf{j}+\cos t \mathbf{k}\right) d t $$

Short Answer

Expert verified
The indefinite integral of the given function is \( e^{t} \mathbf{i} - \cos t \mathbf{j} + \sin t \mathbf{k} + C \)

Step by step solution

01

Break into Components

Break the integral function into its individual components. Each component is related to a unit vector, \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \). These are three separate integrals that need to be calculated independently.
02

Integral of first Component

The first component is \( \int e^{t} dt \). The function \( e^{t} \) when integrated with respect to \( t \), gives \( e^{t} \). Hence, the integrated component corresponding to \( \mathbf{i} \) is \( e^{t} \mathbf{i} \).
03

Integral of second component

The second component is \( \int \sin t dt \). The function \( \sin t \) gives \( -\cos t \) upon integration. Therefore, the component corresponding to \( \mathbf{j} \) would be \( -\cos t \mathbf{j} \).
04

Integral of third Component

The third component is \( \int \cos t dt \). The integral of \( \cos t \) is \( \sin t \). Thus, the integrated component related to \( \mathbf{k} \) is \( \sin t \mathbf{k} \).
05

Sum of components

Combine all the components together to get the final answer. Don't forget to add the constant of integration, \( C \), to the solution after performing the integration.
06

Write the final answer

The final indefinite integral of the given vector-valued function is therefore \( e^{t} \mathbf{i} - \cos t \mathbf{j} + \sin t \mathbf{k} + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Components
Integration by parts is a powerful integration technique used when dealing with products of functions. While we might think it is directly linked to this problem, the method used here involves breaking down a vector-valued function into simpler components. Each part is integrated separately.
This isn't the classic integration by parts method which relies on the formula:
  • \(\int u\, dv = uv - \int v\, du\)
Instead, our task deals with integrating each component individually. The key takeaway is to understand how each function behaves under integration.
  • The exponential function \(e^t\) integrates to itself: \(\int e^{t} dt = e^{t}\).
  • The trigonometric functions \(\sin t\) and \(\cos t\) integrate to \(-\cos t\) and \(\sin t\) respectively.
Therefore, integration by components is not about using integration by parts directly, but breaking down complex functions into smaller integrals for easy computation.
Vector-Valued Functions
Vector-valued functions assign a vector to each element in their domain. In the context of this exercise, we have the function \( e^{t} \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k} \). Each component ties to a direction in the 3-dimensional space represented by \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).
Here's how you understand them:
  • \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are unit vectors pointing in the direction of x, y, and z axes respectively.
  • The function value at each \( t \) is a vector made up of three scalar components.
When integrating a vector-valued function, we conduct the integration for each individual component:
  • The x-component (\( e^{t} \mathbf{i} \)) moves along the \(\mathbf{i}\) direction.
  • The y-component (\( \sin t \mathbf{j} \)) is governed by the \(\mathbf{j}\) direction.
  • The z-component (\( \cos t \mathbf{k} \)) follows the \(\mathbf{k}\) direction.
After integrating each component independently, we combine them to get the resultant vector function.
Constant of Integration
In the context of indefinite integrals, the constant of integration \( C \) arises due to the nature of antiderivatives. When performing an indefinite integration, the set of all antiderivatives of a function includes a constant term. This term accounts for any constant that, when differentiated, results in zero.
Therefore:
  • After integrating a function like \( f(x) \), the result is \( F(x) + C \).
  • This signifies that there are infinitely many antiderivatives for any given function, distinguished only by a constant difference.
In our vector-valued function integration, \( C \) becomes crucial:
  • Each scalar component, when integrated, has its own constant of integration.
  • For the final vector, it unifies as a singular constant \( C \), representing a vector of constants for each dimension.
Remembering to include \( C \) in your solution ensures completeness and correctness of any indefinite integral problem.

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Most popular questions from this chapter

Sketch the graph of the plane curve given by the vector-valued function, and, at the point on the curve determined by \(\mathbf{r}\left(t_{0}\right)\), sketch the vectors \(\mathbf{T}\) and \(\mathbf{N}\). Note that \(\mathbf{N}\) points toward the concave side of the curve. $$ \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j} \quad t_{0}=\frac{\pi}{4} $$

Find the curvature \(K\) of the plane curve at the given value of the parameter. \(\mathbf{r}(t)=5 \cos t \mathbf{i}+4 \sin t \mathbf{j}, \quad t=\frac{\pi}{3}\)

Describe the motion of a particle if the tangential component of acceleration is \(0 .\)

Find the curvature \(K\) of the curve, where \(s\) is the arc length parameter. $$ \mathbf{r}(s)=\left(1+\frac{\sqrt{2}}{2} s\right) \mathbf{i}+\left(1-\frac{\sqrt{2}}{2} s\right) \mathbf{j} $$

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The velocity vector points in the direction of motion.
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