Chapter 14: Problem 5
Evaluate the iterated integrals. $$ \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} d y d x $$
Short Answer
Expert verified
The value of the iterated integral is 2.
Step by step solution
01
Understand the Problem
We are asked to evaluate the iterated integral \( \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} \ dy \ dx \). This involves integrating with respect to \(y\) first and then \(x\).
02
Integrate with Respect to y
First, integrate \( e^{x+y} \) with respect to \(y\). The antiderivative of \( e^{x+y} \) with respect to \(y\) is \( e^{x+y} \). Evaluate this from \(y = 0\) to \(y = \ln 2\): \[\int_{0}^{\ln 2} e^{x+y} \ dy = \left[ e^{x+y} \right]_{0}^{\ln 2} = e^{x+\ln 2} - e^{x}.\]
03
Rewrite the Integral
After integrating with respect to \(y\), the iterated integral becomes: \[\int_{0}^{\ln 3} \left( e^{x+\ln 2} - e^{x} \right) \ dx.\]
04
Simplify the Expression
Simplify the expression \( e^{x+\ln 2} \). Recall that \( e^{x+\ln 2} = e^{x} \cdot e^{\ln 2} = 2e^{x} \). Therefore, the integral is now: \[\int_{0}^{\ln 3} \left( 2e^{x} - e^{x} \right) \ dx = \int_{0}^{\ln 3} e^{x} \ dx.\]
05
Integrate with Respect to x
Now integrate \( e^{x} \) with respect to \(x\). The antiderivative of \( e^{x} \) is \( e^{x} \). Evaluate this from \(x = 0\) to \(x = \ln 3\): \[\int_{0}^{\ln 3} e^{x} \ dx = \left[ e^{x} \right]_{0}^{\ln 3} = e^{\ln 3} - e^{0} = 3 - 1 = 2.\]
06
Conclude the Solution
After evaluating both integrals, we find that the value of the iterated integral is \(2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral calculates the net area under a curve between two specified bounds, which are typically denoted as the lower and upper limits of integration. In simple terms, it provides the accumulation or total of quantities over a given interval. For example, in the iterated integral \[ \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} \, dy \, dx \]we are finding the total accumulation of the function \( e^{x+y} \) over the region defined by the given limits. When evaluating a definite integral, we first need to find the antiderivative of the function with respect to one variable, and then evaluate it using the given limits. Here's the step-by-step process:
- Integrate: Find the antiderivative of the function with respect to the inner variable, which in this case is \( y \).
- Evaluate: Substitute the limits of the inner integral and subtract to find the total accumulation along that dimension.
- Repeat: Use the result to perform the integration with respect to the next variable \( x \).
Exponential Functions
Exponential functions are mathematical expressions that contain a constant base raised to a variable exponent, typically of the form \( f(x) = a^x \) or \( e^x \). In calculus, the natural exponential function \( e^x \) is especially significant due to its unique properties, including its simple derivative and integral. In our exercise, the function \( e^{x+y} \) is a product of exponential terms involving two variables. This feature allows us to separate and simplify expressions during integration, as illustrated below:
- Multiplicative Identity: Exponential terms can be broken apart using properties such as \( e^{x+y} = e^x \cdot e^y \).
- Simplification: This property enables us to simplify \( e^{x + \ln 2} \) as \( 2e^x \). The expression \( e^{\ln 2} = 2 \) since \( e^{\ln a} = a \).
- Calculus Operations: The derivative and integral of \( e^x \) are both \( e^x \), making it straightforward to integrate functions involving \( e^x \).
Antiderivative
The antiderivative, often called the indefinite integral, is the reverse operation of differentiation. It is a function that, when differentiated, gives the original function back. For instance, if \( F(x) \) is an antiderivative of \( f(x) \), then \( F'(x) = f(x) \). In the context of the iterated integral problem, we need to find antiderivatives to solve definite integrals.### Antiderivative of \( e^{x+y} \) with respect to \( y \)Here, the antiderivative of \( e^{x+y} \) with respect to \( y \) is still \( e^{x+y} \), as differentiation of \( e^{x+y} \) with respect to \( y \) would simply reproduce the same function. Here's why:
- When taking the integral with respect to \( y \), consider \( x \) as a constant, thus the integral operates just as \( e^{y} \).
- After integration, applying the limits of \( y = 0 \) to \( y = \ln 2 \) results in the expression \( e^{x + \ln 2} - e^{x} \).