Chapter 14: Problem 2
Evaluate the iterated integral. $$\int_{0}^{2} \int_{-1}^{1}(x-y) d y d x$$
Short Answer
Expert verified
The iterated integral evaluates to 4.
Step by step solution
01
Understand the Double Integral
The given problem involves evaluating the iterated integral \( \int_{0}^{2} \int_{-1}^{1} (x-y)\, dy \, dx \). The integral is iterated, which means we integrate the function \( f(x, y) = x-y \) first with respect to \( y \) and then with respect to \( x \).
02
Integrate with Respect to y
First, we need to find the inner integral with respect to \( y \):\[ \int_{-1}^{1} (x-y) \, dy. \]Integrating term by term, we have:\[ \int (x-y) \, dy = xy - \frac{y^2}{2} \bigg|_{-1}^{1}. \]
03
Evaluate the Inner Integral
Now substitute the limits \(-1\) and \(1\) into the result of the inner integral:\[xy - \frac{y^2}{2} \bigg|_{-1}^{1} = \left( x(1) - \frac{1^2}{2} \right) - \left( x(-1) - \frac{(-1)^2}{2} \right). \]Simplifying:\[ = (x - \frac{1}{2}) - (-x - \frac{1}{2}) = x - \frac{1}{2} + x + \frac{1}{2} = 2x. \]
04
Integrate with Respect to x
Next, we find the integral with respect to \( x \):\[ \int_{0}^{2} 2x \, dx. \]Calculate the integral:\[ \int 2x \, dx = x^2 \bigg|_{0}^{2}. \]
05
Evaluate the Outer Integral
Finally, substitute the limits \(0\) and \(2\) into the result:\[ x^2 \bigg|_{0}^{2} = 2^2 - 0^2 = 4 - 0 = 4. \]
06
Write the Final Answer
The value of the iterated integral \( \int_{0}^{2} \int_{-1}^{1} (x-y) \, dy \, dx \) is 4.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Iterated Integrals
Iterated integrals play a crucial role in multivariable calculus as they allow us to solve integrals over regions in two or more dimensions.They are called "iterated" because we perform integration in a step-by-step manner, one variable at a time.In the given exercise, we deal with the iterated integral \( \int_{0}^{2} \int_{-1}^{1} (x-y) \, dy \, dx \).
To solve this, we first tackle the inner integral \( \int_{-1}^{1} (x-y) \, dy \), integrating with respect to \( y \).After finding this result, we then move on to integrate the outer integral with respect to \( x \).This process is methodical and ensures that each variable's effect is considered individually, providing a clear picture of the function's behavior over the specified region.
To solve this, we first tackle the inner integral \( \int_{-1}^{1} (x-y) \, dy \), integrating with respect to \( y \).After finding this result, we then move on to integrate the outer integral with respect to \( x \).This process is methodical and ensures that each variable's effect is considered individually, providing a clear picture of the function's behavior over the specified region.
- First integrate over the inner variable \( y \).
- Next, substitute the limits for the inner integral.
- Then, integrate over the outer variable \( x \).
Integration Techniques
Integration techniques are tools to solve integrals efficiently and accurately. In the context of iterated integrals, understanding how to break down a complex integral into simpler components is essential. For the inner integral \( \int (x-y) \, dy \), we apply basic integration rules. The function \( x-y \) can be separated and integrated term by term:
- \( \int x \, dy = xy \) since \( x \) is treated as a constant.- \( \int -y \, dy = -\frac{y^2}{2} \) using the power rule of integration.
Once we integrate and substitute the limits \( y = 1 \) to \( y = -1 \), we get a function in \( x \): \( 2x \).This result becomes the new integrand for the outer integral.The outer integral \( \int 2x \, dx \) is straightforward, and we apply similar integration rules:The integration process is simplified by breaking it into manageable steps, showing how each part of the original function contributes to the overall solution.
- \( \int x \, dy = xy \) since \( x \) is treated as a constant.- \( \int -y \, dy = -\frac{y^2}{2} \) using the power rule of integration.
Once we integrate and substitute the limits \( y = 1 \) to \( y = -1 \), we get a function in \( x \): \( 2x \).This result becomes the new integrand for the outer integral.The outer integral \( \int 2x \, dx \) is straightforward, and we apply similar integration rules:
- \( \int 2x \, dx = x^2 \), again using the power rule.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. This allows for a more comprehensive understanding of how variables interact over a region.In the exercise, the function \( f(x, y) = x-y \) represents a two-variable function that we integrate over a specific area, bounded by the limits for \( x \) and \( y \).
Multivariable calculus involves concepts such as partial derivatives and multiple integrals, like integrals over two-dimensional spaces. The solution of this exercise demonstrates a basic application of these concepts.
Multivariable calculus involves concepts such as partial derivatives and multiple integrals, like integrals over two-dimensional spaces. The solution of this exercise demonstrates a basic application of these concepts.
- The inner integral \( \int_{-1}^{1} \) focuses on one-dimensional integration over \( y \).
- The outer integral \( \int_{0}^{2} \) takes the integrated result and finishes the multidimensional integration over \( x \).