Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![### Evaluating an Iterated Integral
**Problem Statement:**
Sketch the region \( R \) and evaluate the iterated integral \( \int_R \int f(x, y) \, dA \).
**Integral Expression:**
\[
\int_0^1 \int_0^2 (1 - 2x + 8y) \, dy \, dx
\]
**Graph Analysis:**
1. **Top Left Graph:**
- This is a shaded rectangular region in the first quadrant.
- The rectangle extends from \( x = 0 \) to \( x = 1 \) and from \( y = 0 \) to \( y = 2 \).
- The filled area represents the region of integration for the given iterated integral.
2. **Top Right Graph:**
- This graph depicts a triangular region in the first quadrant.
- The base of the triangle lies on the \( x \)-axis from \( x = 1 \) to \( x = 2 \).
- The height of the triangle extends from \( y = 0 \) to \( y = 1 \).
- This figure is showing a different possible region, not the original intended rectangle.
3. **Bottom Graph:**
- This graph illustrates another triangular region.
- The base of the triangle lies on the line \( x = 0 \) to \( x = 1 \).
- The height of the triangle goes from \( y = 0 \) to \( y = 2 \).
- This figure also represents a different region compared to the original rectangle.
**Conclusion:**
- The integral \( \int_R \int f(x, y) \, dA = \) (value to be calculated).
- The problem demonstrates how to understand and sketch the region \( R \) for evaluating a double integral.
**Note:** The correct sketch conforms to the integral limits \( x = 0 \) to \( x = 1 \) and \( y = 0 \) to \( y = 2 \). The sketches show alternate regions, with only the top left graph accurately depicting \( R \).
Use these insights to calculate the iterated integral over the specified region.](https://content.bartleby.com/qna-images/question/6fa7ac50-e43d-4267-956c-145cd19bd43c/bf6bdf33-56a9-4eea-8943-5771cd16fa09/cki52e4_processed.png)
Transcribed Image Text:### Evaluating an Iterated Integral
**Problem Statement:**
Sketch the region \( R \) and evaluate the iterated integral \( \int_R \int f(x, y) \, dA \).
**Integral Expression:**
\[
\int_0^1 \int_0^2 (1 - 2x + 8y) \, dy \, dx
\]
**Graph Analysis:**
1. **Top Left Graph:**
- This is a shaded rectangular region in the first quadrant.
- The rectangle extends from \( x = 0 \) to \( x = 1 \) and from \( y = 0 \) to \( y = 2 \).
- The filled area represents the region of integration for the given iterated integral.
2. **Top Right Graph:**
- This graph depicts a triangular region in the first quadrant.
- The base of the triangle lies on the \( x \)-axis from \( x = 1 \) to \( x = 2 \).
- The height of the triangle extends from \( y = 0 \) to \( y = 1 \).
- This figure is showing a different possible region, not the original intended rectangle.
3. **Bottom Graph:**
- This graph illustrates another triangular region.
- The base of the triangle lies on the line \( x = 0 \) to \( x = 1 \).
- The height of the triangle goes from \( y = 0 \) to \( y = 2 \).
- This figure also represents a different region compared to the original rectangle.
**Conclusion:**
- The integral \( \int_R \int f(x, y) \, dA = \) (value to be calculated).
- The problem demonstrates how to understand and sketch the region \( R \) for evaluating a double integral.
**Note:** The correct sketch conforms to the integral limits \( x = 0 \) to \( x = 1 \) and \( y = 0 \) to \( y = 2 \). The sketches show alternate regions, with only the top left graph accurately depicting \( R \).
Use these insights to calculate the iterated integral over the specified region.
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