Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 34
Find the area \(A\) of the region in the first quadrant bounded by the curves \(y=3 e^{x}\) and \(y=2+e^{2 x}\).
Short Answer
Expert verified
The area is \(2.25 - 2\\\ln(2)\).
Step by step solution
01
Find Intersection Points
Set the equations equal to each other to find where the curves intersect: \( 3e^{x} = 2 + e^{2x} \). Rearrange it to \( e^{2x} - 3e^{x} + 2 = 0 \). Let \( u = e^{x} \), then the equation becomes \( u^{2} - 3u + 2 = 0 \). Factor it to get \((u - 1)(u - 2) = 0\), giving solutions \(u = 1\) and \(u = 2\). Convert back to \(x\) by setting \(e^{x} = 1\) or \(e^{x} = 2\), resulting in intersection points \(x = 0\) and \(x = \ln(2)\).
02
Set Up Integral for Area
The area between the two curves can be found using the integral of the difference of the functions. The area \(A\) is given by \( \int_{0}^{\ln(2)} (3e^{x} - (2 + e^{2x})) \, dx \). Simplify the integrand to get \(3e^{x} - 2 - e^{2x}\).
03
Evaluate the Integral
Split the integral into separate integrals: \( \int_{0}^{\ln(2)} 3e^{x} \, dx - \int_{0}^{\ln(2)} 2 \, dx - \int_{0}^{\ln(2)} e^{2x} \, dx \). Evaluate each integral: - The first integral is \( 3 \int_{0}^{\ln(2)} e^{x} \, dx = 3[e^{x}]_{0}^{\ln(2)} = 3(e^{\ln(2)} - e^{0}) = 3(2 - 1) = 3 \).- The second integral is \( 2 \int_{0}^{\ln(2)} 1 \, dx = 2[x]_{0}^{\ln(2)} = 2(\ln(2) - 0) = 2\ln(2) \).- The third integral is \( \frac{1}{2} \int_{0}^{\ln(2)} e^{2x} \, dx = \frac{1}{2}[\frac{1}{2}e^{2x}]_{0}^{\ln(2)} = \frac{1}{2}(\frac{1}{2}e^{2x}) \Big|_{0}^{\ln(2)} = \frac{1}{4}(4 - 1) = \frac{3}{4} \).
04
Compute the Total Area
Add the results of the integrals to find the final area: \[ A = \left( 3 - 2\ln(2) - \frac{3}{4} \right) = \left( 3 - 2\ln(2) - 0.75 \right) = 2.25 - 2\ln(2) \].
05
Interpret the Result
The area between the curves is \( 2.25 - 2\ln(2) \), which represents the space in the first quadrant where the values of \( y = 3e^{x} \) are above \( y = 2 + e^{2x} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Integral Calculus
Integral calculus is a fundamental concept in mathematics. It's used to calculate the area under a curve on a graph. Think of it as adding up all the tiny pieces of area beneath a curve. This helps us find the total space between two different curves. The process of integration allows us to sum these "slices," which can be thin rectangles, to get an accurate area measurement.
When we calculate the area between curves, it's important to set up an integral with limits. These limits define where we start and stop our calculation. In the solved exercise, the limits are from 0 to \(\ln(2)\), thanks to the intersection points of the curves. By setting up our integral this way, we account for the region where one curve is consistently above the other. Hence, the area between them can be calculated correctly. Integral calculus provides the tools we need to work with these kinds of problems.
When we calculate the area between curves, it's important to set up an integral with limits. These limits define where we start and stop our calculation. In the solved exercise, the limits are from 0 to \(\ln(2)\), thanks to the intersection points of the curves. By setting up our integral this way, we account for the region where one curve is consistently above the other. Hence, the area between them can be calculated correctly. Integral calculus provides the tools we need to work with these kinds of problems.
Utilizing Intersection Points
Intersection points are where two graphs meet on a coordinate plane. These points are crucial in determining the limits of integration when finding the area between curves. To find them, we set the two function equations equal to one another.
In our example, we equated \(3e^x\) and \(2 + e^{2x}\). By solving \(3e^x = 2 + e^{2x}\), we discovered the intersection occurs at \(x = 0\) and \(x = \ln(2)\). These points tell us the boundaries of the area shared by the two curves.
Identifying where the curves intersect ensures that the calculated area really represents the space they enclose, starting and ending exactly at these important points. This technique is a key step in the process because it accurately defines the region to be analyzed.
In our example, we equated \(3e^x\) and \(2 + e^{2x}\). By solving \(3e^x = 2 + e^{2x}\), we discovered the intersection occurs at \(x = 0\) and \(x = \ln(2)\). These points tell us the boundaries of the area shared by the two curves.
Identifying where the curves intersect ensures that the calculated area really represents the space they enclose, starting and ending exactly at these important points. This technique is a key step in the process because it accurately defines the region to be analyzed.
Working with Exponential Functions
Exponential functions often appear in integral calculus problems, as they describe a rapid rate of change. They are expressed as \(y = ae^{kx}\), where \(a\) and \(k\) are constants. These functions grow or decay quickly, which makes them interesting to work with.
In the original exercise, we dealt with two exponential functions: \(y = 3e^x\) and \(y = 2 + e^{2x}\). The interplay between these functions was key to determining the area we calculated. As \(x\) increases, \(e^x\) becomes much larger, impacting the area between the curves significantly.
Solving an exponential equation often involves letting \(e^x\) equal a simpler variable like \(u\), which allows us to handle it with algebraic methods. After solutions are found, converting back to terms of \(x\) gives us the actual points of intersection and thus the limits of integration for our area calculation. This step highlights the blend of algebra and calculus in tackling exponential functions, making it manageable and straightforward to understand.
In the original exercise, we dealt with two exponential functions: \(y = 3e^x\) and \(y = 2 + e^{2x}\). The interplay between these functions was key to determining the area we calculated. As \(x\) increases, \(e^x\) becomes much larger, impacting the area between the curves significantly.
Solving an exponential equation often involves letting \(e^x\) equal a simpler variable like \(u\), which allows us to handle it with algebraic methods. After solutions are found, converting back to terms of \(x\) gives us the actual points of intersection and thus the limits of integration for our area calculation. This step highlights the blend of algebra and calculus in tackling exponential functions, making it manageable and straightforward to understand.
