Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. \(\int_{0}^{\pi / 2} \int_{0}^{1} x \sin x y d y d x\)

Step-by-Step Solution

Answer

The iterated integral evaluates to \( \frac{\pi}{2} - 1 \).

Step 1: Understand the Problem

We are given a double integral \( \int_{0}^{\pi / 2} \int_{0}^{1} x \sin(xy) \ dy \ dx \). The outer integral is with respect to \( x \) from 0 to \( \pi/2 \), while the inner integral is with respect to \( y \) from 0 to 1. We'll tackle this by first evaluating the inner integral.

Step 2: Evaluate the Inner Integral

Focus on the inner integral \( \int_{0}^{1} x \sin(xy) \ dy \). Treat \( x \) as a constant. The substitution \( u = xy \) implies \( du = x \, dy \), so \( dy = \frac{du}{x} \). Change the limits: when \( y = 0, u = 0 \) and when \( y = 1, u = x \). The integral becomes \( \int_{0}^{x} \sin u \ du \).

Step 3: Integrate with Respect to u

Now, integrate \( \sin u \) with respect to \( u \):\[\int \sin u \ du = -\cos u + C\]Apply the limits from 0 to \( x \):\[\left[-\cos(u)\right]_{0}^{x} = -\cos(x) + \cos(0) = 1 - \cos(x)\]

Step 4: Evaluate the Outer Integral

Substitute the result from the inner integral into the outer integral:\[\int_{0}^{\pi / 2} \left(1 - \cos(x)\right) \ dx\]Split it into two integrals:\[\int_{0}^{\pi / 2} 1 \, dx - \int_{0}^{\pi / 2} \cos(x) \, dx\]

Step 5: Solve the Separated Integrals

The first integral is straightforward:\[\int_{0}^{\pi / 2} 1 \, dx = \left[x\right]_{0}^{\pi / 2} = \frac{\pi}{2}\]For the second integral:\[\int_{0}^{\pi / 2} \cos(x) \, dx = \left[\sin(x)\right]_{0}^{\pi / 2} = 1 - 0 = 1\]

Step 6: Combine Results

Subtract the result of the cosine integral from the first integral:\[\frac{\pi}{2} - 1\]

Step 7: Conclusion

The value of the iterated integrals is \( \frac{\pi}{2} - 1 \).

Key Concepts

Iterated Integration

Double integrals like the one in the exercise involve performing multiple integrations in a specific order. For iterated integration, we process one integral inside another, dependent on the limits provided. The order of integration is critical, as it determines the boundaries over which each variable is integrated. In our problem, the outer integral involves the variable \( x \), and ranges from 0 to \( \pi/2 \), while the inner integral encompasses the variable \( y \), spanning 0 to 1. The procedure often involves evaluating the innermost integral first, treating any outer variables as constants, and then proceeding outwardly. This approach follows an organized path, step by step, through each layer of integration.

Integration by Substitution

The substitution method is a vital technique in solving integrals, including those encountered in double integrals. In our specific problem, the substitution \( u = xy \) is made within the inner integral. By treating \( x \) as a constant, the substitution aids in transforming a complex integral into a simpler form.Here's how it works:

• Define the substitution: \( u = xy \), which gives us \( du = x \, dy \).
• Recalculate the limits of integration: When \( y = 0, u = 0 \) ; when \( y = 1, u = x \).
• Transform the integral: This substitution converts the integral of \( x \sin(xy) \) with respect to \( y \) into an integral of \( \sin(u) \) with respect to \( u \).

This technique significantly simplifies the integral, allowing us to carry out the calculations in a straightforward manner.

Trigonometric Integrals

Trigonometric integrals commonly appear in calculus, and handling them often requires specific methods. In our case, the resulting integral after substitution involved \( \sin(u) \). The process of integrating trigonometric functions, such as \( \sin(u) \), involves fundamental integration rules:

• The integral of \( \sin(u) \) is \( -\cos(u) + C \).
• Apply the limits to this antiderivative: convert back from \( u \) limits to actual numerical boundaries.

Understanding how to handle basic trigonometric integrals is essential as they form the core of many calculus problems.

Calculus Problem Solving

Tackling calculus problems requires a systematic approach, especially in multi-step processes like double integrals.Think of problem-solving in calculus as a puzzle involving:

• Breaking down the problem: Identify each part of the integral and the methods needed.
• Applying techniques: Use integration methods like substitution or trigonometric integration as required.
• Checking each step: Ensure each part, whether substitution, limits, or constants, is correctly applied.

By methodically walking through each layer of complexity in calculus problems, you ensure a smooth and logical resolution, leading to the correct outcome, much like reaching \( \frac{\pi}{2} - 1 \) in our practice solution.