Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. $$ \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} d y d x $$

Step-by-Step Solution

Answer

The value of the integral is 2.

Step 1: Understand the Structure of the Iterated Integral

The given integral is \( \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} \, dy \, dx \). This is an iterated integral where we first integrate with respect to \( y \) from \( y = 0 \) to \( y = \ln 2 \), and then integrate the result with respect to \( x \) from \( x = 0 \) to \( x = \ln 3 \). Our task is to solve the inner integral first, then the outer integral.

Step 2: Evaluate the Inner Integral with Respect to y

The inner integral is \( \int_{0}^{\ln 2} e^{x+y} \, dy \). We can rewrite the integrand as \( e^x \cdot e^y \). Since \( e^x \) is constant with respect to \( y \), it can be treated as a constant during integration.\[\int_{0}^{\ln 2} e^{x+y} \, dy = e^x \int_{0}^{\ln 2} e^y \, dy\]The integral \( \int e^y \, dy = e^y + C \) can be evaluated from \( 0 \) to \( \ln 2 \).

Step 3: Solve the Inner Integral

Continuing from the equation: \[e^x \left[ e^y \right]_{0}^{\ln 2} = e^x \left( e^{\ln 2} - e^0 \right) = e^x \left(2 - 1\right) = e^x \]So, the result of the inner integral is \( e^x \).

Step 4: Evaluate the Outer Integral with Respect to x

Now, evaluate the outer integral using the result from the inner integral:\[\int_{0}^{\ln 3} e^x \, dx\]The integral of \( e^x \) is \( e^x \). Evaluate it from \( 0 \) to \( \ln 3 \).

Step 5: Solve the Outer Integral

Using the antiderivative of \( e^x \), we find:\[\left[ e^x \right]_{0}^{\ln 3} = e^{\ln 3} - e^0 = 3 - 1 = 2\]Thus, the value of the entire iterated integral is 2.

Key Concepts

Integration Techniques

Integration techniques are key to solving various types of integrals. For iterated integrals, like the one in our problem, it is helpful to think of them as double integrations, where you integrate a function of two variables. You perform the integration one variable at a time. This means starting with the inner integral, which is with respect to one variable, in our case, \( y \). Once this is solved, the result is treated as a regular integral with respect to the outer variable, \( x \).

• Understand how the order of integration—whether starting with \( x \) or \( y \)—may affect the result if limits change.
• Recognize constants during integration, which can simplify the work, as seen when \( e^x \) is treated as a constant while integrating \( e^{x+y} \) with respect to \( y \).

Practicing different approaches and becoming familiar with these steps can help you tackle more complex integrals confidently.

Exponential Functions

Exponential functions, such as \( e^{x+y} \), are characterized by the base \( e \), an important constant roughly equal to 2.71828. These functions have effective properties that make them common in integration problems. Let's look into some key aspects:- **Constant Factor:** When integrating exponentials with variables in the exponent, observe constants or splits, like when \( e^{x+y} \) becomes \( e^x \cdot e^y \).- **Antiderivative Knowledge:** Recognize that the integral of \( e^u \), where \( u \) is a variable or a linear expression, remains \( e^u \), which simplifies many integrations.In the given exercise, we use these exponential properties to simplify solving the integral, managing the constant \( e^x \) separately while focusing on the simpler integral of \( e^y \).

Definite Integrals

Definite integrals are those with specified upper and lower limits, giving a numerical result rather than a general form. They are about calculating the net area between the curve and the x-axis over a given interval. In the context of an iterated integral, - **Step-by-Step Evaluation:** Each definite integral is first evaluated fully within the inner limits before moving on to the outer integral with its own limits.- **Plug in Limits:** After finding the antiderivative, calculate the value by plugging in the upper limit and subtracting the result from what you get with the lower limit. For example, in our integral, the solution involves \( e^{\ln 2} \) and \( e^0 \).This step-by-step evaluation ensures precise calculations and effectively uses the properties of definite integrals to arrive at a specific solution, as was done to obtain the final solution of 2 in our example.