Problem 14 - Thomas Calculus

Question

In Exercises \(1-14,\) evaluate the iterated integral. $$\int_{-1}^{2} \int_{1}^{2} x \ln y d y d x$$

Step-by-Step Solution

Answer

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

Step 1: Understand the Iterated Integral

We need to evaluate the iterated integral given by \( \int_{-1}^{2} \int_{1}^{2} x \ln y\, dy\, dx \). This means we will integrate with respect to \( y \) first, and then with respect to \( x \).

Step 2: Integrate with Respect to y

The inner integral is \( \int_{1}^{2} x \ln y\, dy \). Since \( x \) is treated as a constant with respect to \( y \), we can factor it out of the integral. We then integrate \( \ln y \) with respect to \( y \), which gives the antiderivative \( y \ln y - y \). Thus, the inner integral becomes:\[x \left[ y \ln y - y \right]_{1}^{2}\]

Step 3: Evaluate the Inner Integral

Substitute the limits of integration for \( y \) into the antiderivative:\[x \left[ (2 \ln 2 - 2) - (1 \ln 1 - 1) \right] = x \left( 2 \ln 2 - 2 + 1 \right) = x (2 \ln 2 - 1)\]Thus, the result of the inner integral is \( x (2 \ln 2 - 1) \).

Step 4: Integrate with Respect to x

Now, evaluate the outer integral:\[\int_{-1}^{2} x (2 \ln 2 - 1)\, dx\]Since \( 2 \ln 2 - 1 \) is a constant, we can factor it out of the integral:\[(2 \ln 2 - 1) \int_{-1}^{2} x\, dx\]

Step 5: Evaluate the Outer Integral

Find the antiderivative of \( x \) which is \( \frac{x^2}{2} \). Substitute the limits of integration for \( x \):\[(2 \ln 2 - 1) \left[ \frac{x^2}{2} \right]_{-1}^{2} = (2 \ln 2 - 1) \left( \frac{2^2}{2} - \frac{(-1)^2}{2} \right)\]Simplify:\[= (2 \ln 2 - 1) \left( \frac{4}{2} - \frac{1}{2} \right) = (2 \ln 2 - 1) \cdot \frac{3}{2}\]

Step 6: Calculate the Final Result

Multiply out the expression to get the final result:\[\frac{3}{2} (2 \ln 2 - 1) = 3 \ln 2 - \frac{3}{2}\]The evaluated value of the integral is \( 3 \ln 2 - \frac{3}{2} \).

Key Concepts

Integration by Parts

Integration by Parts is a technique used to integrate products of functions. It's derived from the product rule for differentiation. The formula is:

• \[ \int u \, dv = uv - \int v \, du \]

To apply this method, you choose one function to differentiate (\(u\)) and one to integrate (\(dv\)). This is useful when the product of functions involves algebraic terms multiplied by logarithmic or exponential functions. However, in iterated integrals like the one we are dealing with, Integration by Parts is not directly used because the focus is primarily on integrating one variable at a time step by step, rather than manipulating the integrals themselves. In our example, the integration is straightforward due to how the functions work together. Nonetheless, knowing this technique is beneficial when more complex products of functions appear.

Antiderivative

An antiderivative, also called an indefinite integral, represents a function whose derivative is the original function. For instance, we seek the antiderivative when we are asked to integrate \( \ln y \) with respect to \( y \), leading to \( y \ln y - y \). Here are some common examples of antiderivatives:

• The antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) if \( n eq -1 \).
• The antiderivative of \( e^x \) is \( e^x \).
• The antiderivative of \( \ln x \) is \( x \ln x - x \).

In iterated integrals, finding the correct antiderivative is crucial for evaluating each integral individually. When integrating over multiple variables, it helps simplify each step before moving on to the next variable. This method is essential for accurately finding the solution, as demonstrated when simplifying \( 2 \ln 2 - 1 \) for our problem.

Limits of Integration

Limits of Integration define the interval over which an integral is evaluated. They mark the boundaries along which you accumulate the area or volume under the curve or surface. When dealing with iterated integrals, keep these points in mind:

• Evaluate the inner integral first within its defined limits before proceeding to the outer integral.
• Changing the order of integration may require different limits.

In our example, the limits are clearly stated as \([-1, 2]\) for \(x\) and \([1, 2]\) for \(y\). This straightforward setup means you work through each integrand sequentially without complex adjustments. Paying close attention to these limits helps ensure the process is correct, leading to accurate solutions.

Multivariable Calculus

Multivariable Calculus extends calculus concepts to functions of several variables. This field covers several key topics:

• Partial Derivatives: Derivatives concerning one variable while keeping others constant.
• Multiple Integrals: Like iterated integrals, these include both double and triple integrals, allowing evaluations over areas and volumes.
• Vector Calculus: Involves functions with multiple inputs and outputs, often graphed in higher-dimensional spaces.

In iterated integrals, these concepts allow us to evaluate complex functions over specific regions. Our problem specifically demonstrates integration of a two-variable function by first integrating over \(y\), then \(x\), which is a primary method within multivariable calculus. As you master these topics, you gain the tools necessary to explore and solve problems involving complex systems and multidimensional functions.