Problem 11 - University Calculus: Early Transcendentals

Question

Evaluate the iterated integral. $$\int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x d x d y$$

Step-by-Step Solution

Answer

The iterated integral evaluates to \( \frac{3}{2} \).

Step 1: Set Up the Double Integral

The iterated integral is given by \( \int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x \ dx \ dy \). Recognize that we will first integrate with respect to \( x \) while treating \( y \) as a constant.

Step 2: Perform the Inner Integral

Evaluate the inner integral \( \int_{0}^{\pi / 2} y \sin x \ dx \). Since \( y \) is a constant with respect to \( x \), we take it outside of the integration: \( y \int_{0}^{\pi / 2} \sin x \ dx \). The integral of \( \sin x \) is \( -\cos x \), so evaluate from 0 to \( \frac{\pi}{2} \): \( y[ -\cos(\frac{\pi}{2}) + \cos(0) ] = y[ 0 + 1 ] = y \).

Step 3: Integrate the Result with Respect to y

Now evaluate the outer integral: \( \int_{-1}^{2} y \ dy \). The antiderivative of \( y \) is \( \frac{1}{2}y^2 \). Evaluate from -1 to 2: \([\frac{1}{2}(2)^2] - [\frac{1}{2}(-1)^2] = [\frac{1}{2} \cdot 4] - [\frac{1}{2} \cdot 1] = 2 - \frac{1}{2} = \frac{3}{2} \).

Step 4: Conclude the Solution

The value of the iterated integral \( \int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x \ dx \ dy \) is \( \frac{3}{2} \).

Key Concepts

Double Integral

A double integral is an extension of a single integral to functions of two variables. In a double integral, you are integrating a function over a two-dimensional region, such as a rectangle or a more complex shape. For instance, in the original exercise, we evaluated the double integral \[ \int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x \, dx \, dy \]over a rectangular region determined by \(-1 \leq y \leq 2 \) and \(0 \leq x \leq \frac{\pi}{2}\). You typically perform the integration in two steps:

• The inner integral is computed first, integrating with respect to one variable while keeping the other constant.
• The result of the inner integration is then used to perform the outer integral.

Understanding how to set up the limits of integration is crucial, as they define the area over which the function is integrated.

Integration Techniques

Integration techniques are methods used to find antiderivatives or to evaluate integrals. In the context of double integrals, these methods allow us to simplify and effectively compute the required areas. The original exercise involved integrating the function \( y \sin x \) over the specified bounds.For the inner integral \( \int_{0}^{\pi / 2} \sin x \, dx \), we applied basic integration techniques for trigonometric functions. The function \( \sin x \) integrates to \(-\cos x\). Then, by evaluating from the limits 0 to \( \frac{\pi}{2}\), we obtain:

• \( \sin x \) becomes \(-\cos(\frac{\pi}{2}) + \cos(0) = 0 + 1 = 1\).

Following this, we integrate with respect to \( y \):

• Recognize that \( y \) remains constant when integrating with respect to \( x \).
• Take \( y \) as a constant outside the inner integral, simplifying the process.
• Finally, perform the outer integration resulting in the evaluated value for the iterated integral.

Calculus Problem Solving

Solving calculus problems involves understanding the process and being able to apply various techniques efficiently. Here's a structured approach to solving problems like the original exercise:1. **Identify the Region:** Clearly understand the limits of integration and the region over which the function is defined.
2. **Select the Correct Order of Integration:** Decide whether to first integrate with respect to \( x \) or \( y \). This can be crucial if the region or function makes one order more straightforward.
3. **Integrate in Steps:** Break down the problem: handle the inner integral first, then use its result for the outer integral.

• Ensure to apply the fundamental theorem of calculus correctly.
• Simplify at each step to avoid errors and to make subsequent steps easier.

4. **Conclude with the Final Solution:** Sum everything up, check your work, and make sure the final result makes sense contextually, as we did with the iterated integral in the exercise.These strategies not only help in solving typical calculus problems but also in developing a deeper understanding of how integrals work in multi-dimensional spaces.