Chapter 12: Problem 1
Evaluate the integrals. $$\int_{0}^{1}\left[t^{3} \mathbf{i}+7 \mathbf{j}+(t+1) \mathbf{k}\right] d t$$
Short Answer
Expert verified
\( \frac{1}{4} \mathbf{i} + 7 \mathbf{j} + \frac{3}{2} \mathbf{k} \)
Step by step solution
01
Understand the Integral
The given integral is a vector integral, which involves integrating each component of the vector separately. The vector is \( \mathbf{F}(t) = t^3 \mathbf{i} + 7 \mathbf{j} + (t+1) \mathbf{k} \).
02
Integrate the "i" Component
Evaluate the integral of the "i" component. The "i" component is \( t^3 \). Integrate \( \int_{0}^{1} t^3 \, dt \).
03
Solve the "i" Component Integral
Using the power rule for integration, \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \), the integral becomes \( \frac{t^4}{4} \). Evaluate it from 0 to 1: \( \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \).
04
Integrate the "j" Component
The "j" component is a constant, \( 7 \). Integrate \( \int_{0}^{1} 7 \, dt \).
05
Solve the "j" Component Integral
Integrating a constant \( k \) over \( a \) to \( b \) gives \( k(b-a) \). Thus, the integral is \( 7 \times (1 - 0) = 7 \).
06
Integrate the "k" Component
The "k" component is \( t+1 \). Integrate \( \int_{0}^{1} (t+1) \, dt \).
07
Solve the "k" Component Integral
Separate the integral: \( \int t \, dt + \int 1 \, dt \). This equals \( \frac{t^2}{2} + t \). Evaluate from 0 to 1: \( \left( \frac{1^2}{2} + 1 \right) - \left( \frac{0^2}{2} + 0 \right) = \frac{1}{2} + 1 = \frac{3}{2} \).
08
Combine the Results
Combine the evaluated integrals from each component into a vector: \( \frac{1}{4} \mathbf{i} + 7 \mathbf{j} + \frac{3}{2} \mathbf{k} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Integral
A vector integral extends the concept of a typical integral to vector-valued functions. Unlike regular integrals, which deal with scalars, vector integrals focus on vectors composed of several components. Each component is a separate integral. In the presented exercise, the vector function is given as \( \mathbf{F}(t) = t^3 \mathbf{i} + 7 \mathbf{j} + (t+1) \mathbf{k} \).
The final result is a new vector consisting of the integrated components.With vector integrals, keep in mind:
- The \( \mathbf{i} \) component is \( t^3 \)
- The \( \mathbf{j} \) component is \( 7 \)
- The \( \mathbf{k} \) component is \( t+1 \)
The final result is a new vector consisting of the integrated components.With vector integrals, keep in mind:
- Break down the vector into its individual scalar component functions.
- Integrate each component independently within the specified limits.
- Combine the results to form the integrated vector result.
Integration of Vector Functions
Integration of vector functions involves treating each component of a vector function as an independent scalar function. This is a valuable technique in vector calculus as it simplifies complex integrals into manageable parts. Let's illustrate this with the example:
In the vector function \( \mathbf{F}(t) = t^3 \mathbf{i} + 7 \mathbf{j} + (t+1) \mathbf{k} \), each component \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) is integrated separately:
\( \int_{0}^{1}(t^3 \mathbf{i} + 7 \mathbf{j} + (t+1) \mathbf{k}) \, dt = \frac{1}{4} \mathbf{i} + 7 \mathbf{j} + \frac{3}{2} \mathbf{k} \).This approach enables more straightforward calculation and understanding of the integral of vector functions.
In the vector function \( \mathbf{F}(t) = t^3 \mathbf{i} + 7 \mathbf{j} + (t+1) \mathbf{k} \), each component \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) is integrated separately:
- For the \( \mathbf{i} \) component: Integrate \( t^3 \) over the interval \([0,1]\).
- For the \( \mathbf{j} \) component: Integrate the constant \(7\) over \([0,1]\).
- For the \( \mathbf{k} \) component: Integrate \(t+1\) over \([0,1]\).
\( \int_{0}^{1}(t^3 \mathbf{i} + 7 \mathbf{j} + (t+1) \mathbf{k}) \, dt = \frac{1}{4} \mathbf{i} + 7 \mathbf{j} + \frac{3}{2} \mathbf{k} \).This approach enables more straightforward calculation and understanding of the integral of vector functions.
Power Rule for Integration
The power rule is a fundamental integration technique that simplifies finding antiderivatives. It applies to expressions with powers of a variable. For any function \( t^n \), where \( n \) is a constant, the integration is performed as follows:
The rule states:\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]where \( C \) is the constant of integration.
In the given exercise, the integration of \( t^3 \) illustrates the use of the power rule:
The rule states:\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]where \( C \) is the constant of integration.
In the given exercise, the integration of \( t^3 \) illustrates the use of the power rule:
- \( n = 3 \), so \( \int_{0}^{1} t^3 \, dt = \frac{t^4}{4} \)
- Evaluate from 0 to 1: \( \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \)
- Increase the power by one (\( n+1 \)).
- Divide by the new power.
- Apply limits for definite integrals.