Question

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{1}^{2} \int_{-1}^{1}\left(\frac{x}{y}+3\right) d x d y $$

Short answer

(a) 6, (b) \( \int_{-1}^{1} \int_{1}^{2} (\frac{x}{y} + 3) dy \, dx \)

Step-by-step solution

Understand the Iterated Integral

The given integral is \( \int_{1}^{2} \int_{-1}^{1} \left(\frac{x}{y} + 3\right) dx \, dy \). This represents integrating with respect to \(x\) first while \(y\) is held constant, and then integrating the result with respect to \(y\).

Integrate with Respect to x

To perform the inner integral, \( \int_{-1}^{1} \frac{x}{y} + 3 \, dx \), treat \(y\) as a constant:\[ \int_{-1}^{1} \left(\frac{x}{y} + 3\right) dx = \int_{-1}^{1} \frac{x}{y} \, dx + \int_{-1}^{1} 3 \, dx \]Compute each:- The integral of \(\frac{x}{y}\) with respect to \(x\) is \(\frac{x^2}{2y}\).- The integral of 3 with respect to \(x\) is \(3x\).

Evaluate the Inner Integral

Evaluate each term from \(x = -1\) to \(x = 1\):- For \(\frac{x^2}{2y}\): \[ \frac{1^2}{2y} - \frac{(-1)^2}{2y} = \frac{1}{2y} - \frac{1}{2y} = 0 \]- For \(3x\): \[ 3(1) - 3(-1) = 3 + 3 = 6 \]Thus, the inner integral evaluates to 6.

Integrate with Respect to y

Substitute the result from the inner integration back into the outer integral:\[ \int_{1}^{2} 6 \, dy \]Since 6 is constant with respect to \(y\), this becomes:\[ 6 \times (2 - 1) = 6 \]

Verify and Conclude Part (a)

The evaluated iterated integral is 6. The correctness is confirmed by checking the steps and simplifications above.

Rewrite the Integral with Different Order

Initially, the order was \(\int_{1}^{2} \int_{-1}^{1} (\frac{x}{y} + 3) \ dx \ dy \). For the other order, integrate with respect to \(y\) first. Change the limits, where \(y\) goes from \(-1\) to \(1\), and \(x\) goes from \(1\) to \(2\).The new integral is:\[ \int_{-1}^{1} \int_{1}^{2} \left(\frac{x}{y} + 3\right) dy \, dx \]

Recompute with new order (optional)

To solve this, integrate the expression within \( \int_{1}^{2} (\frac{x}{y} + 3) dy \) as done before but with respect to \(y\). Due to symmetry, the results should again verify the total integral value is 6.

Key concepts

Integration Techniques

Integration techniques are essential in calculus, helping us evaluate complex expressions. Iterated integrals involve integrating a function multiple times with respect to different variables.
Understanding how to simplify these expressions is key.

• For iterated integrals, break down the expression into manageable parts.
• Start by identifying the inner integration, which is crucial for simplifying the solution.
• Always carefully handle the variables, as some are treated as constants while integrating with respect to others.

This step-by-step approach ensures that even complicated integrals become tractable. Practice and familiarity with the fundamental integration rules are vital for mastering these techniques.
For example, consider the inner integral \[ \int_{-1}^{1} \left(\frac{x}{y} + 3\right) \, dx \] where you should treat \( y \) as a constant to make it simpler to handle.
Use basic integration rules on simple functions, like polynomials, to gradually solve complex integrals.

Order of Integration

The order of integration in iterated integrals can be crucial. It changes how you approach solving the integral and can simplify computations.
Traditionally, you might be given an integral with a specific order, such as first integrating with respect to \( x \) and then with \( y \). Switching the order may make evaluations easier, especially if one order has easier limits to work with or simpler functions to integrate.

• Check if reordering helps by first considering the limits of integration for each variable.
• Sometimes, reversing the order simplifies the calculation by allowing easier access to more manageable results.

For example, the integral \[ \int_{1}^{2} \int_{-1}^{1} \left(\frac{x}{y} + 3\right) \, dx \, dy \] can be rewritten as \( \int_{-1}^{1} \int_{1}^{2} \left(\frac{x}{y} + 3\right) \, dy \, dx \).
This concept aligns with recognizing potential simplifications in problems by evaluating which variable limits or functions are easier to handle.

Calculus

Calculus is the mathematical study of continuous change. It is foundational for understanding motion, growth, and other dynamic systems. Iterated integrals fall under the integral calculus branch, used for finding areas, volumes, and other quantities that require aggregation.
Understanding calculus starts with its two main parts:

• Differential Calculus: Focuses on changing rates and slopes of curves.
• Integral Calculus: Concerns summation of quantities and calculation of areas under or between curves.

Iterated integrals form the core of many calculus applications, allowing us to perceive more complex dimensions.
When you tackle iterated integrals:

• Stay grounded in calculus foundations, recognizing how each part contributes to the whole.
• Use the principles of linearity and division in integration to dissect and solve complex integrals.

Mastering these concepts allows you to apply calculus effectively across various fields, from physics to economics.