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Magazine Chapter 2: Problem 27
For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the x-axis or y-axis, whichever seems more convenient. $$ y=x e^{x}, y=e^{x}, x=0, \text { and } x=1 $$
Short Answer
Expert verified
The area between the curves is \( e - 1 \).
Step by step solution
01
Understand the Problem
We need to find the area between the curves given by the equations \( y = x e^x \) and \( y = e^x \) from \( x = 0 \) to \( x = 1 \). This involves graphing these functions, identifying the region between them, and calculating the area using integration.
02
Graph the Functions
Start by sketching the graphs of \( y = x e^x \) and \( y = e^x \) over the interval \( x = 0 \) to \( x = 1 \). \( y = e^x \) is an exponential curve starting at 1 when \( x = 0 \), while \( y = x e^x \) begins at 0 and increases as a product of \( x \) and exponential growth.
03
Determine the Area Boundaries
The region of interest is enclosed by the curves \( y = e^x \) on the top, \( y = x e^x \) on the bottom, and both vertical lines \( x = 0 \) and \( x = 1 \). This defines our limits for integration.
04
Set Up the Integral
Since the curves are given as functions of \( x \), it's most straightforward to integrate with respect to \( x \). The area \( A \) is given by integrating the difference \( e^x - x e^x \) from \( x = 0 \) to \( x = 1 \): \[ A = \int_0^1 (e^x - x e^x) \, dx \]
05
Evaluate the Integral
Use integration techniques to solve the integral. This involves computing: \[ \int (e^x - x e^x) \, dx = \int e^x \, dx - \int x e^x \, dx \]The first term is straightforward: \[ \int e^x \, dx = e^x \]For the second term, apply integration by parts where \( u = x \) and \( dv = e^x \, dx \): \[ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x \]Thus, \[ A = \left[ e^x - (x e^x - e^x) \right]_0^1 = \left[ e^x - x e^x + e^x \right]_0^1 \]
06
Compute the Definite Integral
Evaluate the expression from \(0\) to \(1\):\[ \left[ e^x - x e^x + e^x \right]_0^1 = \left[ 2e^x - x e^x \right]_0^1 \]\[ = \left( (2e - 1e) - (2\cdot1 - 0) \right) = (2e - e - 1) = e - 1\]
07
Conclude with Area
The area of the region between the curves \( y=x e^x \) and \( y=e^x \), from \( x = 0 \) to \( x = 1 \), is \( e - 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Functions
Graphing functions helps us visually understand relations and trends between variables. Seeing a function on a graph provides insight into its behavior and allows us to identify features like intersections, turning points, and areas of interest. When graphing functions like \( y = x e^x \) and \( y = e^x \), begin by plotting key points along the specified interval, in this case, from \( x = 0 \) to \( x = 1 \).
- \( y = e^x \) is a basic exponential function that starts at 1 when \( x = 0 \) and steedily increases as \( x \) grows.
- \( y = x e^x \) combines linear and exponential growth, starting at 0 and increasing on the same interval.
Integration Techniques
Integration is a fundamental calculus tool for calculating areas, volumes, and other quantities. It involves finding the total size or value by summing infinitesimally small parts of a function.
There are several integration techniques, such as:
There are several integration techniques, such as:
- Basic integration rules, e.g., \( \int e^x \, dx = e^x + C \)
- Integration by substitution, useful for handling composite functions
- Integration by parts, a technique applying the integration of products of functions based on the formula \( \int u \, dv = uv - \int v \, du \)
Integration by Parts
Integration by parts is a powerful method derived from the product rule of differentiation. It is particularly useful for products of functions, such as in our exercise when integrating \( x e^x \).
The process involves choosing:
The process involves choosing:
- \( u \), which will be differentiated
- \( dv \), which will be integrated
Calculating Area Between Curves
Calculating the area between curves in calculus involves integrating the difference of two functions over a specific interval. In our exercise, the functions are \( y = e^x \) and \( y = x e^x \).
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) is:\[ A = \int_0^1 (e^x - x e^x) \, dx \]Once the integral is set up, break it down:
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) is:\[ A = \int_0^1 (e^x - x e^x) \, dx \]Once the integral is set up, break it down:
- Evaluate \( \int e^x \, dx = e^x \) straightforwardly.
- Use integration by parts for \( \int x e^x \, dx \), as described earlier.
