Problem 13 - University Calculus: Early Transcendentals

Question

Evaluate the iterated integral. $$\int_{1}^{4} \int_{1}^{e} \frac{\ln x}{x y} d x d y

Step-by-Step Solution

Answer

The value of the iterated integral is \( \ln 2 \).

Step 1: Set Up the Integral

We are given the iterated integral \( \int_{1}^{4} \int_{1}^{e} \frac{\ln x}{x y} \, dx \, dy \). The order of integration is \( dx \, dy \). First, recognize that \( x \) is to be integrated over \([1, e]\) and \( y \) over \([1, 4]\). The integral is set up correctly for these bounds.

Step 2: Integrate with Respect to \( x \)

The inner integral is \( \int_{1}^{e} \frac{\ln x}{x y} \, dx \). Notice that \( \frac{1}{y} \) is a constant with respect to \( x \), so it can be taken outside the integral: \( \frac{1}{y} \int_{1}^{e} \frac{\ln x}{x} \, dx \). The integral \( \int \frac{\ln x}{x} \, dx \) can be solved using substitution, with \( u = \ln x \), \( du = \frac{1}{x} \, dx \). The antiderivative of \( u \, du \) is \( \frac{u^2}{2} + C \). This becomes \( \frac{(\ln x)^2}{2} \).

Step 3: Evaluate the Inner Integral

Now, we evaluate the antiderivative of the inner integral from 1 to \( e \):\[ \frac{1}{y} \left[ \frac{(\ln x)^2}{2} \right]_{1}^{e} = \frac{1}{y} \left( \frac{(\ln e)^2}{2} - \frac{(\ln 1)^2}{2} \right) = \frac{1}{y} \left( \frac{1}{2} - 0 \right) = \frac{1}{2y} \].

Step 4: Integrate with Respect to \( y \)

Now substitute back into the remaining integral: \( \int_{1}^{4} \frac{1}{2y} \, dy \). This can be rewritten as \( \frac{1}{2} \int_{1}^{4} \frac{1}{y} \, dy \). The integral of \( \frac{1}{y} \, dy \) is \( \ln |y| + C \).

Step 5: Evaluate the Outer Integral

Evaluate \( \frac{1}{2} \left[ \ln |y| \right]_{1}^{4} = \frac{1}{2} \left( \ln 4 - \ln 1 \right) = \frac{1}{2} \ln 4 \). Since \( \ln 1 = 0 \), the expression simplifies to \( \frac{1}{2} \ln 4 = \frac{1}{2} (2 \ln 2) = \ln 2 \).

Key Concepts

Substitution Method

The substitution method is a fundamental technique in calculus for simplifying the process of finding antiderivatives. It's especially handy when the integral isn't straightforward to compute directly. In substitution, we

• Choose a new variable, typically denoted as \( u \), that simplifies the integral.
• Express the original variable in terms of this new variable.
• Transform the differential accordingly.

For example, given the integral \( \int \frac{\ln x}{x} \, dx \), substitution can make the integral more manageable. Here, you might set \( u = \ln x \), which simplifies the integration process. Since derivating \( u \) gives \( du = \frac{1}{x} \, dx \), the integral can now be transformed into \( \int u \, du \), which is easier to evaluate as the antiderivative is \( \frac{u^2}{2} + C \). This method is quite powerful because it replaces complex-looking expressions with simpler ones aligned with basic integration formulas.

Integration by Parts

Integration by parts is another essential technique that helps when substitution alone doesn't make the function easily integrable. It's derived from the product rule for differentiation and follows the formula:\[\int u \, dv = uv - \int v \, du\]In situations where you have a product of functions, choose \( u \) to be a part of the function that becomes simpler when differentiated, and \( dv \) as the part you can directly integrate.

• Identify \( u \) and \( dv \).
• Differiate to get \( du \) and integrate to get \( v \).
• Apply the integration by parts formula.

This technique is especially useful when working with logarithmic or power functions, as it systematically reduces the problem into solvable components. However, for some integrals, a combination of integration by parts and substitution might be necessary for simplification.

Natural Logarithm

The natural logarithm, denoted \( \ln x \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. This function is

• Fundamental in calculus due to its natural occurrence in growth processes, solving integrals, and differential equations.
• The derivative of the natural logarithm \( \ln x \) is \( \frac{1}{x} \), making it deeply useful in integration techniques.
• Appears in problems such as the integral \( \int \frac{1}{y} \, dy = \ln |y| + C \), where the natural logarithm simplifies expressions.

The natural logarithm often appears when performing integration, especially in terms of simplifying integrands or expressing solutions in their simplest forms. Its properties and relations, such as \( \ln 1 = 0 \) and \( \ln e = 1 \), are frequently used in calculations.

Definite Integrals

Definite integrals are used to calculate the exact area under a curve within specified bounds, represented as \( \int_{a}^{b} f(x) \, dx \). Unlike indefinite integrals, which yield a general form with a constant \( + C \), definite integrals result in a numerical value.

• The endpoints \( a \) and \( b \) define the range over which the integration occurs.
• Evaluation involves calculating the antiderivative at \( b \) and \( a \) and finding their difference: \([F(b) - F(a)]\).
• Applied when calculating precise quantities like areas or averages within an interval.

For instance, in the original exercise's outer integral \( \int_{1}^{4} \frac{1}{y} \, dy \), evaluating this leads to finding \( \left[ \ln |y| \right]_{1}^{4} = \ln 4 - \ln 1 \). The definite integral is a powerful tool in solving real-world problems, particularly those involving accumulation, such as finding total area or volume.