Problem 19 - University Calculus: Early Transcendentals

Question

Sketch the region of integration and evaluate the integral. $$\int_{0}^{\pi} \int_{0}^{x} x \sin y d y d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(1 + \frac{\pi^2}{2}\).

Step 1: Understanding the Integral Limits

The given integral is \( \int_{0}^{\pi} \int_{0}^{x} x \sin y \, d y \, d x \). The limits of integration for \( y \) are from \( 0 \) to \( x \) and for \( x \) are from \( 0 \) to \( \pi \). This defines the region in the xy-plane where \( 0 \le y \le x \) and \( 0 \le x \le \pi \).

Step 2: Sketching the Region

Plot the lines \( y = 0 \) and \( y = x \) within the bounds \( x = 0 \) to \( x = \pi \). The region of integration is the area bounded by these lines which is a right triangle with vertices at \( (0,0) \), \( (\pi,0) \), and \( (\pi, \pi) \).

Step 3: Inner Integral Evaluation

Evaluate the inner integral \( \int_{0}^{x} x \sin y \, d y \), where \( x \) is treated as a constant: \[ \int_{0}^{x} x \sin y \, d y = x \left( -\cos y \right) \bigg|_{0}^{x} = x \left( -\cos x + 1 \right). \]

Step 4: Outer Integral Evaluation

Now evaluate the outer integral: \[ \int_{0}^{\pi} x (-\cos x + 1) \, dx. \] This splits into two integrals: \[ \int_{0}^{\pi} -x \cos x \, dx + \int_{0}^{\pi} x \, dx. \]

Step 5: Solving \( \int_{0}^{\pi} -x \cos x \, dx \)

Use integration by parts for \( \int_{0}^{\pi} -x \cos x \, dx \) with \( u = x \) and \( dv = \cos x \, dx \). Then \( du = dx \) and \( v = \sin x \). This yields: \[ \int u \, dv = uv - \int v \, du = -x \sin x \bigg|_{0}^{\pi} + \int \sin x \, dx = 0 + [1 - 0] = 1. \]

Step 6: Solving \( \int_{0}^{\pi} x \, dx \)

The integral \( \int_{0}^{\pi} x \, dx \) is straightforward: \[ \frac{x^2}{2} \bigg|_{0}^{\pi} = \frac{\pi^2}{2}. \]

Step 7: Final Integration Result

Add the results from the two parts of the integral: \[ 1 + \frac{\pi^2}{2}. \] This gives the value of the original double integral.

Key Concepts

Understanding Integration by Parts

Integration by parts is a fundamental technique used for solving integrals involving products of functions. It's based on the product rule for differentiation. This method is particularly useful when you need to integrate the product of two functions, and one of them is easier to differentiate than integrate. The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]Here, you choose two parts of the integral:

• One part is set as \( u \) and the other as \( dv \).
• Then differentiate \( u \) to get \( du \) and integrate \( dv \) to obtain \( v \).

In our exercise, when we evaluated \( \int_{0}^{\pi} -x \cos x \, dx \), we let \( u = x \) and \( dv = \cos x \, dx \). Then, \( du = dx \) and \( v = \sin x \). Applying the integration by parts formula simplified the problem significantly. It's crucial to strategically select \( u \) and \( dv \) to make the integral easier to solve.

Identifying the Region of Integration

The region of integration is an essential concept when dealing with double integrals. In this exercise, the integral is described by limits which indicate the area of interest in the xy-plane.For the given problem, the bounds for \( y \) range from \( 0 \) to \( x \) and for \( x \), from \( 0 \) to \( \pi \). These limits form a right triangle within the plane:

• The line \( y = 0 \) is the x-axis.
• The line \( y = x \) extends diagonally through the plane.
• The vertical line at \( x = \pi \) bounds the triangle on the right.

The vertices of this triangle are located at points \((0, 0)\), \((\pi, 0)\), and \((\pi, \pi)\). Visualizing the region of integration helps in understanding the scope and constraints of the problem and ensures you properly evaluate the integral within this bounded region.

Solving Trigonometric Integrals

Trigonometric integrals are common in calculus, and they often involve functions like sine and cosine. These integrals require specific techniques and sometimes strategic substitutions or manipulations.In the exercise, we needed to solve \( \int_{0}^{x} x \sin y \, dy \). As \( x \) is treated as a constant here, integration with respect to \( y \) proceeds with a focus on the sine function. The integral of \( \sin y \) is \(-\cos y\), leading to:\[ x \left( -\cos y \right) \bigg|_{0}^{x} = x \left( -\cos x + 1 \right). \]Later, in the double integration process, when handling \( \int_{0}^{\pi} -x \cos x \, dx \), the cosine function is involved. Utilizing integration by parts allows us to manage the cosine function smoothly. Understanding these trigonometric integrals and how to handle them is essential for solving complex integral problems effectively.