Chapter 8: Problem 99
Use a graphing utility to graph the region bounded by the graphs of the equations, and find the area of the region. $$ y=x e^{-x}, y=0, x=4 $$
Short Answer
Expert verified
The area of the region bounded by given equations is \( 4 - 5e^{-4} \) square units.
Step by step solution
01
Graph the Equations
Using a graphing utility, graph each of the equations. The graph of the equation \( y = x e^{-x} \) will be an exponentially decreasing curve. The line \( y = 0 \) is just the x-axis, and \( x = 4 \) is a vertical line.
02
Identify the Region
Observe the plotted graph and identify the region bounded by these three equations. It will be beneath the curve of \( y = x e^{-x} \), above the x-axis (because of \( y = 0 \)), and to the left of the vertical line \( x = 4 \). The region's left boundary is the y-axis.
03
Integrate to Find Area
Use definite integration to compute area of the region under the curve from 0 to 4. The integral which gives the area is \[ \int_{0}^{4} x e^{-x} dx \].
04
Solve The Integral
Applying Integration by parts (where \( u = x \) and \( dv = e^{-x} dx \) ) or apply a graphing calculus utility tool, it can be solved to equal to 4 - 5e^{-4}.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integration
Definite integration is a fundamental concept in calculus used to calculate the area under a curve over a specified interval on the x-axis. In the context of our exercise, we are interested in the area under the curve defined by the equation \( y = x e^{-x} \), bounded by the y-axis (\( x = 0 \)), another vertical line (\( x = 4 \)), and above the x-axis (\( y = 0 \)).
To find this area, we use definite integration, which involves calculating the integral of a function with specific upper and lower limits. These limits, 0 and 4 in this case, represent the range of x-values over which we want to find the area.
To find this area, we use definite integration, which involves calculating the integral of a function with specific upper and lower limits. These limits, 0 and 4 in this case, represent the range of x-values over which we want to find the area.
- The lower limit \(a\) is the point where integration begins (\(x = 0\) for our example).
- The upper limit \(b\) is where integration ends (\(x = 4\) in our scenario).
Exponential Functions
Exponential functions are mathematical functions of the form \( f(x) = a \cdot e^{bx} \), where \( e \) is Euler's number, a constant approximately equal to 2.71828. In our exercise, the function \( y = x e^{-x} \) is an example of an exponential function that decreases as \( x \) increases.
Exponential functions have distinct characteristics:
Exponential functions have distinct characteristics:
- They can model growth or decay processes, like population expansion or radioactive decay.
- The base, \( e \), allows the function to grow or shrink at a rate proportional to its value, leading to continuous growth or decay.
Integration by Parts
Integration by parts is a technique in calculus used to integrate products of functions. It's particularly useful when faced with the integral of two functions multiplied together, as is the case in our problem: \( \int x e^{-x} \, dx \).
The formula for integration by parts is derived from the product rule of differentiation and is expressed as:
\[ \int u \, dv = uv - \int v \, du\]
The formula for integration by parts is derived from the product rule of differentiation and is expressed as:
\[ \int u \, dv = uv - \int v \, du\]
- Choose one function to differentiate (let's call it \( u \)) and another to integrate (\( dv \)).
- Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
- Apply the integration by parts formula to solve for the integral.
- Differentiate \( u \) to get \( du = dx \).
- Integrate \( dv \) to find \( v = -e^{-x} \).