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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![### Calculating the Integral by Interchanging the Order of Integration
Given the integral:
\[ \int_{-1}^{1} \left( \int_{-2}^{2} (2x + 3y + 5) \, dx \right) \, dy \]
We need to interchange the order of integration to solve the integral.
To understand this better:
- The original integration is done first with respect to \(x\) from \(-2\) to \(2\), and then with respect to \(y\) from \(-1\) to \(1\).
To change the order of integration for the region defined:
- The limits for \(x\) and \(y\) need to be reconsidered according to the region of integration.
Let's rewrite the integral with the order of integration changed:
\[ \int_{-2}^{2} \left( \int_{-1}^{1} (2x + 3y + 5) \, dy \right) \, dx \]
This changes the problem such that we first integrate with respect to \(y\) and then with respect to \(x\).
### Steps to Solve the Integral
1. **Inner Integral (with respect to \(y\))**:
\[\int_{-1}^{1} (2x + 3y + 5) \, dy \]
2. **Outer Integral (with respect to \(x\))**:
\[\int_{-2}^{2} \left[ \text{Result of Inner Integral} \right] \, dx \]
#### Integration Steps:
1. Evaluate the inner integral:
\[\int (2x + 3y + 5) \, dy = 2xy + \frac{3y^2}{2} + 5y + C\]
Applying the limits \(-1\) to \(1\):
\[
\left[2xy + \frac{3y^2}{2} + 5y \right]_{-1}^{1} = \left(2x(1) + \frac{3(1)^2}{2} + 5(1)\right) - \left(2x(-1) + \frac{3(-1)^2}{2} + 5(-1)\right)
\]
Simplifying:
\[
\left(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad35cd49-88f1-4759-9e4c-232c0c792d3e%2Fd858beb8-5cdf-4d92-bc4c-daf386bcd4fc%2Fdz55wv.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating the Integral by Interchanging the Order of Integration
Given the integral:
\[ \int_{-1}^{1} \left( \int_{-2}^{2} (2x + 3y + 5) \, dx \right) \, dy \]
We need to interchange the order of integration to solve the integral.
To understand this better:
- The original integration is done first with respect to \(x\) from \(-2\) to \(2\), and then with respect to \(y\) from \(-1\) to \(1\).
To change the order of integration for the region defined:
- The limits for \(x\) and \(y\) need to be reconsidered according to the region of integration.
Let's rewrite the integral with the order of integration changed:
\[ \int_{-2}^{2} \left( \int_{-1}^{1} (2x + 3y + 5) \, dy \right) \, dx \]
This changes the problem such that we first integrate with respect to \(y\) and then with respect to \(x\).
### Steps to Solve the Integral
1. **Inner Integral (with respect to \(y\))**:
\[\int_{-1}^{1} (2x + 3y + 5) \, dy \]
2. **Outer Integral (with respect to \(x\))**:
\[\int_{-2}^{2} \left[ \text{Result of Inner Integral} \right] \, dx \]
#### Integration Steps:
1. Evaluate the inner integral:
\[\int (2x + 3y + 5) \, dy = 2xy + \frac{3y^2}{2} + 5y + C\]
Applying the limits \(-1\) to \(1\):
\[
\left[2xy + \frac{3y^2}{2} + 5y \right]_{-1}^{1} = \left(2x(1) + \frac{3(1)^2}{2} + 5(1)\right) - \left(2x(-1) + \frac{3(-1)^2}{2} + 5(-1)\right)
\]
Simplifying:
\[
\left(
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