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Magazine Chapter 3: Problem 78
Show that $$ \int_{0}^{1} \int_{0}^{y}(y-x)^{n} d x d y=\frac{1}{(n+1)(n+2)} $$
Short Answer
Expert verified
The double integral evaluates to \( \frac{1}{(n+1)(n+2)} \) after integrating over \( x \) and \( y \).
Step by step solution
01
Understand the Problem
We need to evaluate the double integral \( \int_{0}^{1} \int_{0}^{y} (y-x)^{n} \, d x \, d y \) and show that it equals \( \frac{1}{(n+1)(n+2)} \).
02
Integrate with Respect to x
Focus on the inner integral \( \int_{0}^{y} (y-x)^{n} \, dx \). Let \( u = y - x \), so that \( du = -dx \), which changes the limits of integration from \( x=0 \) to \( x=y \), resulting in integration with respect to \( u \): \( \int_{y}^{0} u^n (-du) = \int_{0}^{y} u^n du \). Integrate this to get \( \frac{y^{n+1}}{n+1} \).
03
Substitute Back and Simplify
Since \( u = y - x \) returns to \( x = 0 \) gives \( u = y \) and \( x = y \) gives \( u = 0 \), the substitution works correctly. Therefore, the result for the inner integral is \( \frac{y^{n+1}}{n+1} \).
04
Integrate with Respect to y
Integrate the result of the inner integral: \( \int_{0}^{1} \frac{y^{n+1}}{n+1} \, dy \). This is now a regular single-variable integral which becomes \( \frac{1}{n+1} \cdot \frac{y^{n+2}}{n+2} \bigg|_{0}^{1} \).
05
Evaluate the Final Integral
Evaluate \( \frac{1}{n+1} \cdot \frac{y^{n+2}}{n+2} \bigg|_{0}^{1} = \frac{1}{n+1} \cdot \frac{1}{n+2} = \frac{1}{(n+1)(n+2)} \). The calculation yields the desired result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Change of Variables
The technique of 'change of variables' is a powerful tool in integration that simplifies complex integrals by transforming them into a more easily integrable form. In this exercise, we apply a specific transformation to the original integral, changing the variable from \( x \) to \( u \) where \( u = y - x \). This substitution transforms the integrand \( (y-x)^n \) to \( u^n \), making it much simpler to handle.
- The differential \( dx \) also changes according to the transformation: it becomes \(-du\), due to the derivative \( du = -dx \).
- One must also adjust the limits of integration, which change from \( x=0 \) to \( x=y \), becoming \( u = y \) to \( u = 0 \).
Iterated Integration
Iterated integration refers to the process of integrating a function of two or more variables in sequence, often seen in the double integrals. This is effectively what we do when handling the problem above.
- The function \( (y-x)^n \) is initially integrated with respect to one variable, \( x \), while treating the other, \( y \), as a constant.
- After completing the first integration, we integrate the resulting expression with respect to the remaining variable, \( y \).
Integration Limits
Setting correct integration limits is vital in solving any integral, especially in double integrals. In the given exercise, the integration limits for the inner integral go from \( x = 0 \) to \( x = y \), where \( y \) serves as the upper limit of \( x \).
- These limits define a region of integration in the xy-plane, specifically a triangular region where \( 0 \leq x \leq y \) and \( 0 \leq y \leq 1 \).
- When we change variables and integrate with respect to \( u \) and \( y \), it's crucial to adjust our understanding of these limits accordingly, which is shown by changing from \( 0 \) to \( y \) to \( y \) to \( 0 \).
Polynomial Integration
Polynomial integration involves integrating expressions that feature powers of variables. This process is central in solving the given double integral. As the function \( (y-x)^n \) turns into \( u^n \) after substitution, we perform polynomial integration.
- The integral of \( u^n \) with respect to \( u \) is given by \( \frac{u^{n+1}}{n+1} \) plus a constant of integration, which simplifies the computation.
- After integrating with respect to \( x \) and returning to \( y \), we then integrate a single-variable polynomial, \( y^{n+1} \), with respect to \( y \).
