Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem: Finding the Indefinite Integral**
Find the indefinite integral. (Use \( c \) for the constant of integration.)
\[
\int \left( 4t^3\mathbf{i} + 14t\mathbf{j} - 7 \sqrt{t}\mathbf{k} \right) dt
\]
**Solution:**
To solve the given integral, we need to integrate each component with respect to \( t \):
1. Integrate \( 4t^3 \mathbf{i} \):
\[
\int 4t^3 dt = t^4 + c_1\mathbf{i}
\]
2. Integrate \( 14t \mathbf{j} \):
\[
\int 14t dt = 7t^2 + c_2\mathbf{j}
\]
3. Integrate \( -7 \sqrt{t} \mathbf{k} \):
\[
\int -7 \sqrt{t} dt = -7 \cdot \frac{2}{3} t^{3/2} + c_3 \mathbf{k} = -\frac{14}{3} t^{3/2} + c_3 \mathbf{k}
\]
Combining these results, we get:
\[
\int \left( 4t^3\mathbf{i} + 14t\mathbf{j} - 7 \sqrt{t}\mathbf{k} \right) dt = \left( t^4 + c_1 \right) \mathbf{i} + \left( 7t^2 + c_2 \right) \mathbf{j} - \left( \frac{14}{3} t^{3/2} + c_3 \right) \mathbf{k}
\]
Since \( c_1, c_2, c_3 \) are constants of integration, they can be combined into a single constant vector \( \mathbf{C} = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k} \). Thus, the final answer is:
\[
\left( t^4 \mathbf{i} + 7t^2 \mathbf{j} - \frac{14}{3} t^{3/2} \mathbf{k} \right) + \mathbf{C}
\]
Here, \( \mathbf{C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F985611a2-1702-4cb0-96ad-1209ea311b1f%2F1cfc3706-dcbe-45e7-89a4-36c18a4b145d%2Fnmvfifo_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Finding the Indefinite Integral**
Find the indefinite integral. (Use \( c \) for the constant of integration.)
\[
\int \left( 4t^3\mathbf{i} + 14t\mathbf{j} - 7 \sqrt{t}\mathbf{k} \right) dt
\]
**Solution:**
To solve the given integral, we need to integrate each component with respect to \( t \):
1. Integrate \( 4t^3 \mathbf{i} \):
\[
\int 4t^3 dt = t^4 + c_1\mathbf{i}
\]
2. Integrate \( 14t \mathbf{j} \):
\[
\int 14t dt = 7t^2 + c_2\mathbf{j}
\]
3. Integrate \( -7 \sqrt{t} \mathbf{k} \):
\[
\int -7 \sqrt{t} dt = -7 \cdot \frac{2}{3} t^{3/2} + c_3 \mathbf{k} = -\frac{14}{3} t^{3/2} + c_3 \mathbf{k}
\]
Combining these results, we get:
\[
\int \left( 4t^3\mathbf{i} + 14t\mathbf{j} - 7 \sqrt{t}\mathbf{k} \right) dt = \left( t^4 + c_1 \right) \mathbf{i} + \left( 7t^2 + c_2 \right) \mathbf{j} - \left( \frac{14}{3} t^{3/2} + c_3 \right) \mathbf{k}
\]
Since \( c_1, c_2, c_3 \) are constants of integration, they can be combined into a single constant vector \( \mathbf{C} = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k} \). Thus, the final answer is:
\[
\left( t^4 \mathbf{i} + 7t^2 \mathbf{j} - \frac{14}{3} t^{3/2} \mathbf{k} \right) + \mathbf{C}
\]
Here, \( \mathbf{C
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