Problem 19 - Thomas Calculus

Question

In Exercises \(19-24,\) sketch the region of integration and evaluate the integral. \begin{equation} \int_{0}^{\pi} \int_{0}^{x} x \sin y d y d x \end{equation}

Step-by-Step Solution

Answer

The value of the integral is \( \pi(\pi - 1) \).

Step 1: Understand the Region of Integration

The integral \( \int_{0}^{\pi} \int_{0}^{x} x \sin y \, dy \, dx \) involves two variables: \(x\) and \(y\). The limits for \(y\) are from \(0\) to \(x\), meaning \(y\) ranges from the x-axis up to the line \(y = x\). The limits for \(x\) are from \(0\) to \(\pi\), describing a triangular region in the \(xy\)-plane with vertices at \((0,0)\), \((\pi, 0)\), and \((\pi, \pi)\).

Step 2: Switching the Order of Integration

To evaluate the integral more easily, consider changing the order of integration. The new integral will be \( \int_{0}^{\pi} \int_{y}^{\pi} x \sin y \, dx \, dy \). This new region keeps the same area but switches the roles of \(x\) and \(y\).

Step 3: Evaluate the Inner Integral

Now consider the inner integral \( \int_{y}^{\pi} x \sin y \, dx \). Since \( \sin y \) is treated as a constant with respect to \(x\), the integral simplifies to \[ \sin y \int_{y}^{\pi} x \, dx = \sin y \left[ \frac{x^2}{2} \right]_{y}^{\pi} = \sin y \left( \frac{\pi^2}{2} - \frac{y^2}{2} \right) \].

Step 4: Evaluate the Outer Integral

Substitute the result back into the outer integral: \( \int_{0}^{\pi} \sin y \left( \frac{\pi^2}{2} - \frac{y^2}{2} \right) \, dy \). This splits into two parts: \( \int_{0}^{\pi} \frac{\pi^2}{2} \sin y \, dy - \int_{0}^{\pi} \frac{y^2}{2} \sin y \, dy \).

Step 5: Calculate the First Part of the Outer Integral

For the integral \( \int_{0}^{\pi} \frac{\pi^2}{2} \sin y \, dy \), compute as \( \frac{\pi^2}{2} [-\cos y]_{0}^{\pi} = \frac{\pi^2}{2} (1 - (-1)) = \pi^2 \).

Step 6: Calculate the Second Part of the Outer Integral

Use integration by parts for \( \int_{0}^{\pi} \frac{y^2}{2} \sin y \, dy \). Let \( u = \frac{y^2}{2} \) and \( dv = \sin y \, dy \). Then \( du = y \, dy \) and \( v = -\cos y \) leading to \[ -\frac{y^2}{2} \cos y \bigg|_0^\pi + \int_{0}^{\pi} y (-\cos y) \, dy \]. Evaluate these and simplify to find this part equals \(\pi\).

Step 7: Combine Results

Combine the results from Steps 5 and 6: the total is \( \pi^2 - \pi = \pi(\pi - 1) \).

Step 8: Final Answer

Thus, the value of the integral is \( \pi(\pi - 1) \).

Key Concepts

Region of Integration

In double integration, the region of integration is the area over which we are evaluating the integral. It's essential to visualize this region to understand what part of the plane we're dealing with.
For the integral \( \int_{0}^{\pi} \int_{0}^{x} x \sin y \, dy \, dx \), the region of integration is a triangular area in the \(xy\)-plane.
Here, the boundaries for \(y\) are from \(0\) to \(x\), indicating that for each fixed \(x\), \(y\) will range from \(0\) to \(x\). The boundaries for \(x\) are from \(0\) to \(\pi\).
This describes a triangle with vertices at \((0, 0)\), \((\pi, 0)\), and \((\pi, \pi)\). Visualizing this can help in understanding how to set limits for both variables when performing integration.

Order of Integration

The order of integration refers to the sequence in which you integrate the functions with respect to different variables. Changing the order can simplify the calculation.
In our problem, we initially have \( \int_{0}^{\pi} \int_{0}^{x} x \sin y \, dy \, dx \), which integrates with respect to \(y\) first.
However, switching the order to \( \int_{0}^{\pi} \int_{y}^{\pi} x \sin y \, dx \, dy \) can make integration simpler.
This adjustment generally involves identifying the integration limits for the new order, keeping in mind the region of integration remains unchanged.
Switching the order can sometimes convert a more complicated integral into one that's straightforward to solve.

Integration by Parts

Integration by parts is a technique used to evaluate more complex integrals, especially when the integration of a product of functions is involved. It's based on the product rule for differentiation.
The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).
For the integral \( \int_{0}^{\pi} \frac{y^2}{2} \sin y \, dy \), we choose \(u = \frac{y^2}{2}\) and \(dv = \sin y \, dy\).
This means \(du = y \, dy\) and \(v = -\cos y\).
Applying these constructs helps to transform an initially challenging integral into terms that are easier to manage and solve, breaking down into simpler parts that we can compute separately.

Trigonometric Integrals

Trigonometric integrals involve integrating expressions containing trigonometric functions like sine, cosine, etc. These often need special methods or substitutions.
In this exercise, we encounter trigonometric functions within our region of integration and during the integration process.
For example, to tackle \( \int \sin y \, dy \), one utilizes the antiderivative \(-\cos y \), recognizing standard trigonometric integrals can simplify certain parts of calculations.
Utilizing properties of symmetry or periodicity in trigonometric functions can also often lead to simplified results, easing the full integration process when dealing with definite integrals over specific intervals.