Question

Exercises 55 and 56, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \int_{-1}^{1} \int_{-2}^{2} y d y d x=\int_{-1}^{1} \int_{-2}^{2} y d x d y $$

Short answer

The statement is true. The two integrals are indeed equal.

Step-by-step solution

Solve the integral on the left side

Find the integral of y ranging from -2 to 2, and then find the integral of the resulting function with respect to x, with the limits from -1 to 1. \[\int_{-1}^{1} \int_{-2}^{2} y d y d x = \int_{-1}^{1} \left[\frac{1}{2}y^2\right]_{-2}^{2} dx= \int_{-1}^{1} (2 - 2) dx = 0\]

Solve the integral on the right side

Find the integral of y ranging from -1 to 1, and then find the integral of the resulting function with respect to y, with the limits from -2 to 2. \[\int_{-1}^{1} \int_{-2}^{2} y d x d y= \int_{-2}^{2} \left[ xy\right]_{-1}^{1} dy = \int_{-2}^{2} (y-(-y)) dy = 0 \]

Compare the results

Compare the two values obtained. The integral of the left side is 0 and the integral on the right side is also 0. Therefore, the initial statement is true.

Key concepts

True or False Questions

True or False questions are often used to test understanding of mathematical concepts. In calculus, these questions require the student to determine the validity of a given statement. For the provided exercise, we need to verify if two double integrals are equal under different integration orders.

These challenges often involve:

• Understanding the properties of integrals and functions.
• Recognizing when the values of integrals change due to order or limits.
• Examining if interchanging the order of integration impacts the solution.

By working through these questions, students can reinforce their grasp on integrals, especially when dealing with multiple integration variables.

Integration Order

The order of integration in a double integral significantly affects the process of finding the solution. In our exercise, one must carefully arrange the nested integrals, as seen in the problem statement.

Removing any ambiguity is essential:

• The inner integral is performed first, corresponding to the limit closest to the function being integrated.
• The outer integral is completed second, based on the limits provided for the entire expression.

Despite the temptation to assume that changing the order would alter results, the solution illustrated that both calculations independently simplify to zero under these specific limits. Hence, in this particular case, integrating from \[-1\] to \[1\] with respect to \[x\] first or starting from \[-2\] to \[2\] with \[y\] does not affect the outcome.

Definite Integrals

Definite integrals calculate the area under a curve within specified limits. For double integrals, this expands to determining volumes under surfaces over a defined region. In the exercise at hand, both are evaluated to ensure they equate to zero after running their respective courses.

Key points about definite integrals include:

• They are finite and bound within given upper and lower limits.
• The outcome over a symmetric interval around zero can result in zero, as positive and negative areas can cancel out.
• In the given scenario, both integrations result in a scenario where computations unfold over symmetric intervals, leading to neutralization.

Students must appreciate how averaging effects occur and realize the ultimate goal is to find definite values across the dimensions of integration.