Problem 14 - University Calculus: Early Transcendentals

Question

Evaluate the iterated integral. $$\int_{-1}^{2} \int_{1}^{2} x \ln y d y d x$$

Step-by-Step Solution

Answer

The value of the integral is \( 3 \ln 2 - \frac{3}{2} \).

Step 1: Understand the Iterated Integral

The given iterated integral is \( \int_{-1}^{2} \int_{1}^{2} x \ln y \, dy \, dx \). This means we first integrate with respect to \( y \), keeping \( x \) constant, and then integrate the result with respect to \( x \).

Step 2: Integrate with respect to y

Start by integrating the inner integral \( \int_{1}^{2} x \ln y \, dy \). Given that \( x \) is constant, the integral becomes \( x \int_{1}^{2} \ln y \, dy \). To solve \( \int \ln y \, dy \), we use integration by parts: Let \( u = \ln y \), \( dv = dy \). Then \( du = \frac{1}{y} dy \) and \( v = y \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Thus, we have \[ \int \ln y \, dy = y \ln y - \int y \frac{1}{y} dy = y \ln y - y + C \]. Now, evaluate this from 1 to 2, giving: \[ \left[ y \ln y - y \right]_{1}^{2} = \left( 2 \ln 2 - 2 \right) - \left( 1 \cdot \ln 1 - 1 \right) = 2 \ln 2 - 2 - (0 - 1) = 2 \ln 2 - 1. \] The result of this integral is \( x(2 \ln 2 - 1) \).

Step 3: Integrate with respect to x

Now, integrate \( \int_{-1}^{2} x(2 \ln 2 - 1) \, dx \). Since \( 2 \ln 2 - 1 \) is a constant, this becomes \( (2 \ln 2 - 1) \int_{-1}^{2} x \, dx \). The integral \( \int_{-1}^{2} x \, dx \) is solved by finding the antiderivative of \( x \), which is \( \frac{x^2}{2} \). Evaluate this from -1 to 2: \[ \left[ \frac{x^2}{2} \right]_{-1}^{2} = \left( \frac{2^2}{2} \right) - \left( \frac{(-1)^2}{2} \right) = \left( \frac{4}{2} \right) - \left( \frac{1}{2} \right) = 2 - \frac{1}{2} = \frac{3}{2}. \] Multiply this result by \( 2 \ln 2 - 1 \): \[ (2 \ln 2 - 1) \cdot \frac{3}{2} = \frac{3}{2} \cdot 2 \ln 2 - \frac{3}{2} = 3 \ln 2 - \frac{3}{2}. \]

Step 4: Provide Final Answer

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

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used in calculus to solve integrals where a simple antiderivative is not easily found. It is based on the product rule for differentiation and can be summarized by the formula:

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

This formula tells us that to integrate the product of two functions, you can rearrange the problem into a simpler integral.
To apply integration by parts, choose which part of your function will be \( u \) and which will be \( dv \).

• \( u \): A function that becomes simpler when differentiated (such as \( \ln y \)).
• \( dv \): The remainder of the function which will be more straightforward to integrate (such as \( dy \)).

In the exercise above, integrating \( \ln y \) with respect to \( y \), we chose:

• \( u = \ln y \) and \( dv = dy \).
• Differentiate \( u \) to get \( du = \frac{1}{y} dy \).
• Integrate \( dv \) to find \( v = y \).

Using the integration by parts formula, you can rewrite the integral, making it easier to solve.

Antiderivative

An antiderivative, also known as an indefinite integral, is the reverse process of finding the derivative. In simpler terms, finding an antiderivative is figuring out a function whose derivative matches a given function.
When you have a definite integral to solve, finding the antiderivative is often the first step. This is because the definite integral is usually evaluated by finding the difference between the antiderivative’s values at the limits of integration.
In the example provided, we needed to find the antiderivative of the function \( x \), which is \( \frac{x^2}{2} \). This antiderivative was then used to evaluate:

• \[ \int_{-1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{2} \]
• By substituting the limits, you compute the exact area under the curve for a given interval.

Finding antiderivatives is how integration reduces complex problems, allowing you to calculate areas and more in calculus.

Definite Integral

A definite integral is a concept in calculus used to find the total sum, area, or net change between two points on a function. Unlike an indefinite integral, it returns a numerical value. These numerical results represent total quantities considering any changes over a given interval.
To compute a definite integral, you evaluate the antiderivative at the upper and lower bounds. The result is the difference between these evaluations.
In our exercise, after finding the antiderivative for \( x \), we evaluated across the interval from -1 to 2:

• \[ \int_{-1}^{2} x(2 \ln 2 - 1) \, dx \]

This expression was simplified using:

• \[ (2 \ln 2 - 1) \int_{-1}^{2} x \, dx = (2 \ln 2 - 1) \left( \frac{3}{2} \right) \]

Giving an exact expression for the total net difference across the interval. Ultimately, definite integrals help in practical applications like calculating areas, volumes, and net changes.