Problem 11 - Calculus and its applications

Question

Evaluate. $$ \int_{0}^{1} \int_{1}^{e^{x}} \frac{1}{y} d y d x $$

Step-by-Step Solution

Answer

The value of the double integral is \( \frac{1}{2} \).

Step 1: Evaluate the Inner Integral

First, we evaluate the inner integral with respect to \( y \), which is \( \int_{1}^{e^x} \frac{1}{y} \, dy \). The integral of \( \frac{1}{y} \) is the natural logarithm of \( y \), or \( \ln|y| \). So, the evaluation gives us \( \ln|e^x| - \ln|1| = \ln(e^x) \). Since \( \ln(e^x) = x \), the result of the inner integral is \( x \).

Step 2: Evaluate the Outer Integral

Now that we have the result of the inner integral as \( x \), we proceed with the outer integral \( \int_{0}^{1} x \, dx \). The integral of \( x \) with respect to \( x \) is \( \frac{x^2}{2} \). Evaluating this from 0 to 1, we get \( \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \).

Key Concepts

Inner Integral

When we work on a double integral, we typically begin by addressing the inner integral first. In this exercise, the inner integral is represented by:

• \( \int_{1}^{e^x} \frac{1}{y} \, dy \)

We focus first on integrating with respect to \( y \) while keeping \( x \) constant. Now, the function \( \frac{1}{y} \) when integrated with respect to \( y \) gives us:

• \( \ln|y| \)

To find the value, we substitute the limits of integration, which are from 1 to \( e^x \). This results in:

• \( \ln|e^x| - \ln|1| \)

Because we know \( \ln(e^x) = x \) and \( \ln(1) = 0 \), the expression simplifies to \( x \). This means the outcome of the inner integral is just \( x \), and we proceed with this value to the outer integral.

Outer Integral

Once the inner integral is evaluated, we use its result to compute the outer integral. We've determined that the outcome of the inner integral is \( x \). Consequently, we are left to solve the outer integral:

• \( \int_{0}^{1} x \, dx \)

This integral requires finding the antiderivative of \( x \) with respect to \( x \), which is:

• \( \frac{x^2}{2} \)

Next, we apply the definite integral limits from 0 to 1. Evaluating,

• \( \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} \)

This simplifies to \( \frac{1}{2} \) as the final result of the entire double integration process. Each step helps us deal with complexity by breaking the problem into more manageable parts. This makes it easier to understand, especially when dealing with functions that involve changes in multiple variables.

Natural Logarithm

The natural logarithm is an important mathematical function, often appearing in calculus and integration, particularly in problems like the one we're solving. Defined as \( \ln(x) \), the natural logarithm is the inverse of the exponential function. When integrating functions like \( \frac{1}{y} \), the natural logarithm becomes particularly useful, as seen in the step where we derived:

• \( \int \frac{1}{y} \, dy = \ln|y| + C \)

The natural logarithm has a simple derivative:

• If \( f(y) = \ln(y) \), then \( f'(y) = \frac{1}{y} \)

In our exercise, we evaluated the natural log at the upper and lower limits of the inner integral principal and noted that \( \ln(e^x) = x \), due to the property that the natural log of \( e \) is 1 when raised to any power of \( x \). Understanding these relationships helps simplify complex problems by leveraging the unique properties of the natural logarithm.