Chapter 15: Problem 24
Let \(f\) be continuous on \([0,1]\) and let \(R\) be the triangular region with vertices \((0,0),(1,0),\) and \((0,1)\) . Show that $$\iint_{R} f(x+y) d A=\int_{0}^{1} u f(u) d u$$
Short Answer
Expert verified
The solution shows that \(\iint_{R} f(x+y) dA = \int_{0}^{1} u f(u) du\) via change of variables.
Step by step solution
01
Define the Region R
The triangular region \(R\) is defined by the vertices \((0,0)\), \((1,0)\), and \((0,1)\). This forms a right triangle in the coordinate plane. The line connecting \((1,0)\) to \((0,1)\) is given by \(x + y = 1\). Thus, the region \(R\) consists of points \((x,y)\) where \(x \geq 0\), \(y \geq 0\), and \(x + y \leq 1\).
02
Set Up the Double Integral
To find \(\iint_{R} f(x+y)\, dA\), we must integrate over the triangular region \(R\). The bounds for \(x\) range from \(0\) to \(1\). For each fixed \(x\), \(y\) ranges from \(0\) to \(1-x\). The double integral is thus \[\int_{0}^{1} \int_{0}^{1-x} f(x+y)\, dy\, dx.\]
03
Change of Variables
Perform a change of variables with \(u = x + y\), which transforms the limits of integration. Note that when \(x = 0\), \(y\) ranges from \(0\) to \(1\). Hence, \(u\) ranges from \(0\) to \(1\). The Jacobian of the transformation from \((x, y)\) to \((x, u)\) is 1, since \(dy = du\). Our transformed integral becomes \[\int_{0}^{1} \int_{0}^{u} f(u)\, dx\, du.\]
04
Integrate with Respect to x
Integrate with respect to \(x\) first. Since \(f(u)\) is independent of \(x\), this simplifies to \[\int_{0}^{1} \int_{0}^{u} f(u)\, dx\, du = \int_{0}^{1} \left[ f(u) x \right]_{0}^{u} du = \int_{0}^{1} u f(u)\, du.\]
05
Final Result
We have shown that \(\iint_{R} f(x+y)\, dA = \int_{0}^{1} u f(u)\, du\), as required, by using the transformation and integrating accordingly.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Change of Variables
The change of variables is a mathematical technique often used in calculus to simplify the process of integration when dealing with complex regions or functions. In the context of a double integral, it allows us to transform the original variables, in this case \(x\) and \(y\), to a new set, such as \(u\) and \(v\), which can simplify the region of integration or the function itself.
In the given problem, we changed from \(x + y\) to a single new variable \(u\). This reduces the complexity by translating the integration problem from one over two variables to one over a single function of another variable. This process involves substituting and rearranging the bounds of integration, making the integration much more straightforward.
In the given problem, we changed from \(x + y\) to a single new variable \(u\). This reduces the complexity by translating the integration problem from one over two variables to one over a single function of another variable. This process involves substituting and rearranging the bounds of integration, making the integration much more straightforward.
- Allows handling of complex surfaces or regions.
- Enables simplification of double integrals into more manageable integrals.
- Changes the variables and limits of integration.
Region of Integration
The region of integration in a double integral specifies the set of points over which we are integrating. In the original exercise, the region \(R\) is described as a triangle with vertices at \((0,0), (1,0), (0,1)\).
- It forms a right triangle in the coordinate plane.
- Bounded by lines \(x = 0\), \(y = 0\), and \(x + y = 1\).
- This simple triangular region helps in understanding the limits of integration for the integral.
Jacobian
The Jacobian is a determinant that is used when changing variables in multiple integrals. It provides the factor by which the area (in two dimensions) or volume (in three dimensions) is scaled when undergoing a coordinate transformation.
In the change of variables where \(u = x + y\), the Jacobian was simply 1 in our exercise. This is because the partial derivatives involved in this transformation resulted in a determinant value of one.
In the change of variables where \(u = x + y\), the Jacobian was simply 1 in our exercise. This is because the partial derivatives involved in this transformation resulted in a determinant value of one.
- Essential for correctly adjusting the integral when variables are changed.
- The Jacobian reflects how elements of area or volume are transformed.
- In this problem, a Jacobian of 1 indicates no stretching or compression occurred during the variable switch.
Continuous Function
A continuous function is one where small changes in the input result in small changes in the output, without sudden jumps or breaks. The function \(f(x+y)\) in our exercise is assumed to be continuous, which is a crucial property for applying calculus operations such as integration.
Continuity ensures that:
Continuity ensures that:
- The function is well-behaved over the region of integration.
- Enables the use of tools like the Fundamental Theorem of Calculus.
- Facilitates the transition of the integral into another form through substitution.