Problem 14 - Thomas Calculus

Question

In Exercises \(11-16,\) integrate \(f\) over the given region. Rectangle \(f(x, y)=y \cos x y\) over the rectangle \(0 \leq x \leq \pi\) \(0 \leq y \leq 1\)

Step-by-Step Solution

Answer

The integral diverges.

Step 1: 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 \( \int_{0}^{\pi} \int_{0}^{1} y \cos(xy) \; dy \; dx \).

Step 2: 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 \( \frac{1}{x}\int_{0}^{x} u \cos(u) \; du \).

Step 3: 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, \(\int u \cos(u) \; du = u \sin(u) - \int \sin(u) \; du = u \sin(u) + \cos(u) + C\).

Step 4: 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 \(\sin(x) + \frac{1}{x}(-1)\).

Step 5: 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.

Key Concepts

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:

• \(0 \leq x \leq \pi\)
• \(0 \leq y \leq 1\)

This means we are examining the behavior of \(f(x, y)\) within these bounds, calculating the collective value throughout this specific area.
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 \(x\).

Understanding the rectangle region is fundamental as it defines the physical space your function covers, guiding the limits of your integrals.

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:

• \(w = u\)
• \(dv = \cos(u) \, du\)

Differentiating and integrating respectively, we achieve:

• \(dw = du\)
• \(v = \sin(u)\)

Plug these into the integration by parts formula:\[\int w \, dv = wv - \int v \, dw\]Thus, we deduce:\[\int u \cos(u) \, du = u \sin(u) + \cos(u) + C\]This technique is vital for breaking down complex integrals into digestible parts. The structure it provides transforms challenging problems into manageable tasks.

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:

• \(\int \cos(u) \, du = \sin(u) + C\)
• \(\int \sin(u) \, du = -\cos(u) + C\)

It is crucial when solving double integrals because trigonometric functions can behave unpredictably over different intervals. Using known integrals of trigonometric functions eases the computation and makes manipulating the expressions simpler.

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:

• 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.

Diverging integrals present a unique challenge in mathematics, and the awareness of their traits, like having terms that go towards infinity, helps in predicting such anomalies during evaluations.