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 iterated integral is 2.

Step 1: Evaluate the Inner Integral

To evaluate the iterated integral, start with the inner integral with respect to \(y\). The inner integral is \(\int_{0}^{\ln 2} e^{x+y} \, dy\). This can be rewritten as \(e^x \int_{0}^{\ln 2} e^y \, dy\) because \(e^x\) is constant with respect to \(y\).

Step 2: Integrate with respect to y

Now integrate \(\int_{0}^{\ln 2} e^y \, dy\). The antiderivative of \(e^y\) is itself, \(e^y\). So, the integral becomes \([e^y]_{0}^{\ln 2} = e^{\ln 2} - e^0 = 2 - 1 = 1\). Now substitute back, the inner integral becomes \(e^x \cdot 1 = e^x\).

Step 3: Evaluate the Outer Integral

Now evaluate the outer integral \(\int_{0}^{\ln 3} e^x \, dx\). The antiderivative of \(e^x\) with respect to \(x\) is \(e^x\). So, calculate \([e^x]_{0}^{\ln 3} = e^{\ln 3} - e^0 = 3 - 1 = 2\).

Step 4: Combine and Conclude

The computed value of the outer integral was \(2\). Since the inner integral reduced to \(e^x\), multiplying the two results gives the final value of the iterated integral as \(2\). Therefore, the result of the iterated integral \(\int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} \, dy \, dx\) is \(2\).

Key Concepts

Definite Integrals

Definite integrals are a core concept in calculus, allowing us to find the total area under a curve between two endpoints. In the exercise, evaluating iterated integrals involves integration with specified limits of integration. The limits tell us "from where to where" we need to find the area. In our given task, the definite integral starts at zero and goes to natural logarithm values.

• The inner integral \(\int_{0}^{\ln 2}\) uses zero to natural log of 2.
• The outer integral \(\int_{0}^{\ln 3}\) uses zero to natural log of 3.

Each definite integral provides a precise value, thanks to these limits. This concept is what differentiates definite integrals from indefinite ones, as it results in an exact numerical value, representing the accumulated quantity, like area or volume, over a specified domain.

Exponential Functions

Exponential functions are widely used in calculus because of their unique properties. The function in the integral, \(e^{x+y}\), is an exponential function where the base is Euler's number \(e\), approximately 2.718. This function grows rapidly and has a crucial feature: its derivative is the same as the function itself.

• When integrating \(e^y\) with respect to \(y\), the antiderivative is simply the exponential function \(e^y\).
• This property makes calculations involving exponential functions simpler, especially in iterated integrals.

By using this property, we seamlessly solve the integration, focusing on the exponent of the base, and solve through straightforward evaluation, dealing with an exponential transform of variables, like \(x + y\).

Evaluation of Integrals

The process of evaluating integrals, especially iterated ones, involves systematically working through each layer of the integral. In this specific exercise, you first solve the inner integral with respect to \(y\), then the outer one concerning \(x\).

• Begin with the inner integral: \(\int_{0}^{\ln 2} e^{x+y} \, dy\), rewriting it as \(e^x \int_{0}^{\ln 2} e^y \, dy\).
• Evaluate the antiderivative of \(e^y\), which remains \(e^y\), and plug in the limits from \(0\) to \(\ln 2\).
• Solve the outer integral \(\int_{0}^{\ln 3} e^x \, dx\), apply the same antiderivative rule, and evaluate it with given boundaries.

Such step-by-step evaluation requires a clear understanding of both the process and the handling of exponential expressions. Each completed integral builds upon the result of the previous, demonstrating the nested nature of iterated integrals.