Chapter 7: Problem 4
Evaluate the double integrals.\(\int_{2}^{3} \int_{-1}^{1}(x+2 y) d y d x\)
Short Answer
Expert verified
5
Step by step solution
01
Understand the Integral
This is a double integral of the function \( x + 2y \) with respect to \( y \) and \( x \). The limits are from \( y = -1 \) to \( y = 1 \) and from \( x = 2 \) to \( x = 3 \).
02
Integrate with Respect to y
First, focus on the inner integral and integrate \( x + 2y \) with respect to \( y \). The integral is: \[ \int_{-1}^{1} (x + 2y) dy \].
03
Perform the Inner Integration
Compute the integral: \[ \int_{-1}^{1} (x + 2y) dy = \int_{-1}^{1} x dy + \int_{-1}^{1} 2y dy \]. Since \( x \) is constant with respect to \( y \), we get \[ x \int_{-1}^{1} dy + 2 \int_{-1}^{1} y dy \].
04
Evaluate the Inner Integrals
The first part is: \[ x \int_{-1}^{1} dy = x[y]_{-1}^{1} = x(1 - (-1)) = 2x \]. The second part is: \[ 2 \int_{-1}^{1} y dy = 2 \[ \frac{y^2}{2} \]_{-1}^{1} = 2 \( \frac{1^2}{2} - \frac{(-1)^2}{2} \) = 2(\frac{1}{2} - \frac{1}{2}) = 0 \]. Combining both parts, we get \[ 2x + 0 = 2x \].
05
Integrate with Respect to x
Now integrate the result of the inner integral, \( 2x \), with respect to \( x \) from 2 to 3: \[ \int_{2}^{3} 2x dx \].
06
Perform the Outer Integration
Compute the integral: \[ \int_{2}^{3} 2x dx = 2 \int_{2}^{3} x dx = 2 \[ \frac{x^2}{2} \]_{2}^{3} = \[ x^2 \]_{2}^{3} = 3^2 - 2^2 = 9 - 4 = 5 \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Multiple Integrals
Multiple integrals expand the concept of single-variable integration to more complex, multi-dimensional functions. The double integral you encountered \(\int_{2}^{3} \int_{-1}^{1}(x+2 y) d y d x\) is a classic example. This involves integrating over a two-dimensional region. Here, the function \(x+2y\) gets integrated first with respect to \(y\) and then with respect to \(x\). Let’s understand them better:
- Inner Integral: Integrate with respect to \(y\) under the limits \(-1 \leq y \leq 1\).
- Outer Integral: Integrate the result with respect to \(x\) under the limits \(2 \leq x \leq 3\).
Integration Techniques
To tackle double integrals, you often need to use specific integration techniques efficiently. Here's a breakdown using the given problem:
1. **Separation of Variables**: Notice how we separated \(x + 2y\) into parts that can be handled independently. We split the integral as follows: \(\int_{-1}^{1} (x + 2y) dy \Rightarrow\ \int_{-1}^{1} x dy + \int_{-1}^{1} 2y dy\).
2. **Constant Factorization**: Since \(x\) is treated as a constant within the inner integral, factoring it out simplifies the process: \(\int_{-1}^{1} x dy \Rightarrow\ x \int_{-1}^{1} dy\).
3. **Evaluating Definite Integrals**: Breaking down:
1. **Separation of Variables**: Notice how we separated \(x + 2y\) into parts that can be handled independently. We split the integral as follows: \(\int_{-1}^{1} (x + 2y) dy \Rightarrow\ \int_{-1}^{1} x dy + \int_{-1}^{1} 2y dy\).
2. **Constant Factorization**: Since \(x\) is treated as a constant within the inner integral, factoring it out simplifies the process: \(\int_{-1}^{1} x dy \Rightarrow\ x \int_{-1}^{1} dy\).
3. **Evaluating Definite Integrals**: Breaking down:
- For \(x \int_{-1}^{1} dy\): By performing the integration, we get the width of the interval \(2x\).
- For \int_{-1}^{1} 2y dy\: Evaluating from \-1\ to \1\ cancels out positive and negative areas, resulting in \0\.
Calculus Applications
Understanding and using double integrals enables you to solve real-world problems. Some key applications include:
- Area Calculation: For complicated regions, integrating over the region gives the total area.
- Volume Determination: Extending to triple integrals can give the volume of a 3D object.
- Physics and Engineering: Compute properties like center of mass, electric charge distribution, and more.