4. (a) Evaluate this integral by hand: dy dr 4r- Show all your steps and give an exact answer. (b) For the integral above, sketch the region of integration, and re-write the integral with the opposite order of integration.
4. (a) Evaluate this integral by hand: dy dr 4r- Show all your steps and give an exact answer. (b) For the integral above, sketch the region of integration, and re-write the integral with the opposite order of integration.
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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![### Calculus Problem: Double Integral Evaluation and Region Sketching
#### Problem Statement:
4. (a) Evaluate this integral by hand:
\[
\int_{1}^{3} \int_{4x-3}^{x^2} \frac{6}{x^2} \, dy \, dx
\]
Show all your steps and give an exact answer.
**Solution:**
*(Step-by-step solution needed here)*
---
(b) For the integral above, sketch the region of integration, and re-write the integral with the opposite order of integration.
**Region of Integration:**
To sketch the region of integration, we observe the limits of the integral:
- \( x \) ranges from 1 to 3.
- For a fixed \( x \), \( y \) ranges from \( 4x-3 \) to \( x^2 \).
The region can be visualized on the xy-plane:
1. Identify the curves:
- \( y = 4x - 3 \)
- \( y = x^2 \)
2. Mark the points where these curves intersect within the given \( x \)-limits, \( x \in [1, 3] \).
**Rewriting the Integral:**
To change the order of integration, we need to express the same region in \( y \)-first terms.
1. Identify the \( y \)-limits:
- The lowest value of \( y \) is when \( x = 1 \) and \( y = 4(1)-3 = 1 \).
- The highest value of \( y \) is when \( x = 3 \) and \( y = \min({
4(3)-3, 3^2}) = 9.
Now rewrite the integral with the new limits:
\[
\int_{1}^{9} \int_{{\text{left boundary}}}^{\text{{right boundary}}} \frac{6}{x^2} \, dx \, dy
\]
Here, to correctly determine the new \( x \)-limits as a function of \( y \):
- Solve \( y=4x-3 \) for \( x \), which gives \( x = \frac{y+3}{4} \).
- Solve \( y = x^2 \) for \( x \), which gives \( x = \sqrt{y} \).
Thus, the integral](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F191441f0-66db-457d-b504-dd20fe1485fa%2Fc8d44c55-40b3-4add-95a7-38d94ac86e13%2Frwoell_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Double Integral Evaluation and Region Sketching
#### Problem Statement:
4. (a) Evaluate this integral by hand:
\[
\int_{1}^{3} \int_{4x-3}^{x^2} \frac{6}{x^2} \, dy \, dx
\]
Show all your steps and give an exact answer.
**Solution:**
*(Step-by-step solution needed here)*
---
(b) For the integral above, sketch the region of integration, and re-write the integral with the opposite order of integration.
**Region of Integration:**
To sketch the region of integration, we observe the limits of the integral:
- \( x \) ranges from 1 to 3.
- For a fixed \( x \), \( y \) ranges from \( 4x-3 \) to \( x^2 \).
The region can be visualized on the xy-plane:
1. Identify the curves:
- \( y = 4x - 3 \)
- \( y = x^2 \)
2. Mark the points where these curves intersect within the given \( x \)-limits, \( x \in [1, 3] \).
**Rewriting the Integral:**
To change the order of integration, we need to express the same region in \( y \)-first terms.
1. Identify the \( y \)-limits:
- The lowest value of \( y \) is when \( x = 1 \) and \( y = 4(1)-3 = 1 \).
- The highest value of \( y \) is when \( x = 3 \) and \( y = \min({
4(3)-3, 3^2}) = 9.
Now rewrite the integral with the new limits:
\[
\int_{1}^{9} \int_{{\text{left boundary}}}^{\text{{right boundary}}} \frac{6}{x^2} \, dx \, dy
\]
Here, to correctly determine the new \( x \)-limits as a function of \( y \):
- Solve \( y=4x-3 \) for \( x \), which gives \( x = \frac{y+3}{4} \).
- Solve \( y = x^2 \) for \( x \), which gives \( x = \sqrt{y} \).
Thus, the integral
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