Problem 5 - Thomas Calculus

Question

In Exercises \(1-10,\) 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 \( \frac{\pi^2}{2} + \pi \).

Step 1: Identify the Limits of Integration

The double integral \( \int_{0}^{\pi} \int_{0}^{x} x \sin y \, dy \, dx \) has the outer integral with respect to \( x \) from \( 0 \) to \( \pi \), and the inner integral with respect to \( y \) from \( 0 \) to \( x \). This indicates that for a given \( x \), \( y \) ranges from \( 0 \) to \( x \).

Step 2: Sketch the Region of Integration

The region of integration is a triangle in the \( xy \)-plane. The line \( y = x \) from \( x = 0 \) to \( x = \pi \) forms the boundary of the region, and \( y \) ranges from \( 0 \) to \( x \). This describes a right triangle with vertices at \((0,0)\), \((\pi, 0)\), and \((\pi, \pi)\).

Step 3: Integrate with Respect to \( y \)

To evaluate the integral, first integrate with respect to \( y \). Compute \( \int_{0}^{x} x \sin y \, dy \). Since \( x \) is treated as a constant here, the integral becomes \( x \left[-\cos y\right]_{0}^{x} = x \left(-\cos x + 1\right) \).

Step 4: Simplify the Inner Integral Result

Simplify the expression obtained from the integration over \( y \): \( x - x \cos x \). This result will form the integrand for the next step.

Step 5: Integrate with Respect to \( x \)

Now integrate the simplified expression with respect to \( x \): \( \int_{0}^{\pi} (x - x \cos x) \, dx \). This separates into two integrals: \( \int_{0}^{\pi} x \, dx \) and \( -\int_{0}^{\pi} x \cos x \, dx \).

Step 6: Evaluate the Integral \( \int_{0}^{\pi} x \, dx \)

Compute \( \int_{0}^{\pi} x \, dx \): the antiderivative is \( \frac{x^2}{2} \). Evaluating it gives \( \frac{\pi^2}{2} \).

Step 7: Evaluate the Integral \( -\int_{0}^{\pi} x \cos x \, dx \)

Use integration by parts with \( u = x \) and \( dv = \cos x \, dx \). This leads to \( du = dx \) and \( v = \sin x \). The formula for integration by parts \( \int u \, dv = uv - \int v \, du \) results in: \( x \sin x - \int \sin x \, dx \).

Step 8: Complete the Integration by Parts

Evaluate \( \int \sin x \, dx = -\cos x \), leading to the overall expression \( x \sin x + \cos x \). Evaluate \( \left[x \sin x + \cos x \right]_{0}^{\pi} \), resulting in \(-\pi\).

Step 9: Combine Results

Combine results of the integrals: \( \frac{\pi^2}{2} - (-\pi) \). Simplifying gives \( \frac{\pi^2}{2} + \pi \).

Step 10: Final Result

The evaluation of the integral \( \int_{0}^{\pi} \int_{0}^{x} x \sin y \, dy \, dx \) yields \( \frac{\pi^2}{2} + \pi \).

Key Concepts

Region of Integration

In double integrals, identifying the region of integration is a crucial step before proceeding with any calculations. Imagine the integration domain as a region in a 2-dimensional plane, specifically the xy-plane. For the given integral \( \int_{0}^{\pi} \int_{0}^{x} x \sin y \, dy \, dx \), the boundaries create a specific area we need to consider.The outer integral runs from 0 to \( \pi \), which indicates that the scope of our x values is within this range. Meanwhile, the inner integral runs from 0 to \( x \), which suggests that for each x-value, y can vary from 0 to the current x-value itself. If we visualize this in the xy-plane, it represents a triangular region.To sketch:

• The line \( y = x \) limits one side of the triangle.
• The x-axis (where \( y = 0 \)) forms the base of the triangle.
• The vertical line at \( x = \pi \) closes off the triangle.

The vertices of this triangle are precisely at coordinates (0,0), (\( \pi, 0 \)), and (\( \pi, \pi \)). Understanding this region is fundamental in setting up the limits for subsequent integrations.

Integration by Parts

Integration by parts is a technique used to integrate the product of two functions. It's based on the formula:\[ \int u \, dv = uv - \int v \, du \]This formula is an integration analog of the product rule for differentiation.In the problem, we encounter a product in the outer integral with respect to \( x \): the integral of \( -x \cos x \). To tackle this, we use integration by parts:

• Set \( u = x \), which implies \( du = dx \).
• Set \( dv = \cos x \, dx \), which leads to \( v = \sin x \).

Applying integration by parts:

• \( uv = x \sin x \)
• Integrate \( v \, du \), which is \( \int \sin x \, dx = -\cos x \).

Putting it all together, the integral becomes:\[ x \sin x - \left( -\cos x \right) = x \sin x + \cos x \]Evaluating at the limits from 0 to \( \pi \), this specific integration part contributes \( -\pi \) to the solution.

Trigonometric Integrals

Trigonometric integrals involve integrals of trigonometric functions such as sine, cosine, tangent, etc. They often require special techniques or identities for simplification.In the given exercise, we encounter the integral of \( \sin y \) with respect to \( y \) in the inner integral. The process involves computing:\[ \int_{0}^{x} x \sin y \, dy \]Here, \( x \) is treated as a constant because we're integrating with respect to \( y \). The integral simplifies using:

• Antiderivative of \( \sin y \) is \( -\cos y \).
• Compute the definite integral as \( x \left[-\cos y \right]_{0}^{x} \).

The result of this integral becomes:\[ x \left(-\cos x + 1\right) \]This simple example highlights the importance of knowing basic trigonometric integral results and forms the basis for more complex integrals in further calculus problems.