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Magazine Chapter 15: Problem 14
In Exercises \(11-16,\) integrate \(f\) over the given region. Rectangle \(f(x, y)=y \cos x y\) over the rectangle \(0 \leq x \leq \pi\) \(0 \leq y \leq 1\)
Short Answer
Expert verified
The integral diverges.
Step by step solution
01
Set Up the Double Integral
The given function is \(f(x, y) = y \cos(xy)\). We need to integrate this function over the rectangle defined by the limits: \(0 \leq x \leq \pi\) and \(0 \leq y \leq 1\). Thus, the double integral to evaluate is \( \int_{0}^{\pi} \int_{0}^{1} y \cos(xy) \; dy \; dx \).
02
Evaluate the Inner Integral
First, we integrate with respect to \(y\) while treating \(x\) as a constant. The integral is \( \int_{0}^{1} y \cos(xy) \; dy \). Use substitution: let \(u = xy\), then \(du = x \, dy\), and hence \(dy = \frac{du}{x}\). The limits change from 0 to \(x\) when \(y=0\) to when \(y=1\). The integral becomes \( \frac{1}{x}\int_{0}^{x} u \cos(u) \; du \).
03
Integrate by Parts
To solve \(\int u \cos(u) \; du\), use integration by parts: let \(w = u\) and \(dv = \cos(u)\, du\). Then \(dw = du\) and \(v = \sin(u)\). By integration by parts, \(\int u \cos(u) \; du = u \sin(u) - \int \sin(u) \; du = u \sin(u) + \cos(u) + C\).
04
Evaluate the Result of the Inner Integral
Substitute back for \(u = xy\) and apply the limits from \(0\) to \(x\) to the expression \(\frac{1}{x}[x \sin(x) + \cos(x) - (0 \cdot \sin(0) + \cos(0))]\), which simplifies to \(\sin(x) + \frac{1}{x}(-1)\).
05
Evaluate the Outer Integral
Now evaluate the integral \(\int_{0}^{\pi} \left( \sin(x) - \frac{1}{x} \right) dx \). The integral \(\int_{0}^{\pi} \sin(x) \; dx\) evaluates to \( -\cos(x) \) from 0 to \(\pi\), resulting in \( \cos(0) - (-\cos(\pi)) = 1 + 1 = 2\). The term \(\int_{0}^{\pi} \frac{1}{x} \; dx\) diverges. Thus, the entire integral is \(2 - \text{undefined}\), indicating it diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rectangle Region
A rectangle region in the context of double integrals refers to a two-dimensional area where integration occurs. This region is defined by specific boundaries for two variables, often labeled as \(x\) and \(y\). In our exercise,
the given function \(f(x, y) = y \cos(xy)\) is integrated over a rectangle defined by the intervals:
When setting up the double integral for a rectangle region,
the given function \(f(x, y) = y \cos(xy)\) is integrated over a rectangle defined by the intervals:
- \(0 \leq x \leq \pi\)
- \(0 \leq y \leq 1\)
When setting up the double integral for a rectangle region,
- We start by integrating with respect to \(y\), keeping \(x\) constant (the inner integral).
- Then we process the outer integral by integrating with respect to \(x\).
Integration by Parts
Integration by Parts is a technique derived from the product rule of differentiation. It's especially helpful when faced with integrals involving products of two functions.
In our exercise, we encounter the integral \(\int u \cos(u) \, du\). Here, we apply Integration by Parts using:
In our exercise, we encounter the integral \(\int u \cos(u) \, du\). Here, we apply Integration by Parts using:
- \(w = u\)
- \(dv = \cos(u) \, du\)
- \(dw = du\)
- \(v = \sin(u)\)
Trigonometric Integration
Trigonometric Integration involves integrals that have trigonometric functions. These could be single functions like \(\cos(x)\) or products involving trigonometric identities.
In the original step-by-step solution, we manage \(\int \sin(u) \, du\) and \(\cos(u)\) using straightforward trigonometric relationships:
In the original step-by-step solution, we manage \(\int \sin(u) \, du\) and \(\cos(u)\) using straightforward trigonometric relationships:
- \(\int \cos(u) \, du = \sin(u) + C\)
- \(\int \sin(u) \, du = -\cos(u) + C\)
Divergence of Integrals
Divergence of Integrals occurs when an integral fails to converge to a finite value. In our case, we see divergence in \(\int_{0}^{\pi} \frac{1}{x} \, dx\).
This integral lacks convergence primarily because \(\frac{1}{x}\) spikes towards infinity as it approaches zero from the positive direction.
In calculations, encountering a divergent integral implies:
This integral lacks convergence primarily because \(\frac{1}{x}\) spikes towards infinity as it approaches zero from the positive direction.
In calculations, encountering a divergent integral implies:
- The total value of the area under the curve they're evaluating becomes infinite.
- The integrand grows too large, preventing a finite sum from being captured.
