Chapter 4: Problem 60
Find the area of the region bounded by the given graphs. $$ y=e^{x}, y=e^{-x}, x=-2 $$
Short Answer
Expert verified
The area is \( 2 - (e^2 + e^{-2}) \).
Step by step solution
01
Identify the Region
We need to find the area bounded by the curves \( y = e^x \) and \( y = e^{-x} \), and the line \( x = -2 \). Visually, this means the area where these curves and line intersect, from \( x = -2 \) to where the curves intersect.
02
Find Points of Intersection
To find the intersection points of \( y = e^x \) and \( y = e^{-x} \), we set them equal: \( e^x = e^{-x} \). Taking natural logs of both sides gives us \( x = 0 \). Hence, they intersect at \( x = 0 \).
03
Set Up the Integral for Area
The area between the curves from \( x = -2 \) to \( x = 0 \) can be found using the integral \( \int_{-2}^{0} (e^x - e^{-x}) \, dx \). Here \( e^x \) is the upper function, and \( e^{-x} \) is the lower function on the interval.
04
Integrate the Functions
Solve the integral \( \int (e^x - e^{-x}) \, dx \). The antiderivative of \( e^x \) is \( e^x \), and for \( e^{-x} \), it's \( -e^{-x} \). So, we have \( e^x + e^{-x} \).
05
Evaluate the Definite Integral
Evaluate \( \left[ e^x + e^{-x} \right]_{-2}^{0} \). First, substitute \( x = 0 \): \( e^0 + e^0 = 2 \). Then \( x = -2 \): \( e^{-2} + e^2 \). Compute the difference: \( (2) - (e^2 + e^{-2}) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is fundamental in calculus, especially when you need to find areas under curves or between curves. In simple terms, a definite integral helps calculate the accumulation of quantities, like areas or net changes, over specific intervals.
For example, when you see an integral set up as \( \int_{-2}^{0} (e^x - e^{-x}) \, dx \), this indicates finding the area between the curves defined by the functions \( e^x \) and \( e^{-x} \) from \( x = -2 \) to \( x = 0 \).
Make sure to follow these steps when working with definite integrals:
For example, when you see an integral set up as \( \int_{-2}^{0} (e^x - e^{-x}) \, dx \), this indicates finding the area between the curves defined by the functions \( e^x \) and \( e^{-x} \) from \( x = -2 \) to \( x = 0 \).
Make sure to follow these steps when working with definite integrals:
- Identify the functions involved and their roles. Determine which is the upper function and which is the lower function over the given interval. In our case, \( e^x \) is above \( e^{-x} \).
- Set up the integral. Subtract the lower function from the upper one inside the integral notation.
- Evaluate the integral. Simplify and compute the antiderivative to find the accumulated quantity.
Exponential Functions
Exponential functions are a vital part of calculus and appear frequently in various scientific and real-world scenarios. Let's explore what sets exponential functions like \( y = e^x \) apart.
Exponential functions have the form \( y = a \cdot e^{bx} \), where \( e \) is the base of the natural logarithm, approximately 2.718. Key characteristics include:
Exponential functions have the form \( y = a \cdot e^{bx} \), where \( e \) is the base of the natural logarithm, approximately 2.718. Key characteristics include:
- They grow or decay at a constant percentage rate. The sign and magnitude of the exponent \( b \) control whether they show growth (if positive) or decay (if negative).
- The function \( y = e^x \) represents exponential growth, making it a powerful model for population growth, radioactive decay, and compound interest.
- The inverse relationship is seen in \( y = e^{-x} \), which signifies exponential decay, shrinking over time.
Area Between Curves
Finding the area between curves can be a fascinating aspect of calculus. This involves calculating the space enclosed by the graphs of different functions. The area is determined by integrating the difference of the functions over a specific interval.
To determine the area between two curves, follow these steps:
To determine the area between two curves, follow these steps:
- Identify the upper and lower functions over the given interval. Typically, this can be discerned visually or mathematically by considering the range of interest; \( y = e^x \) is above \( y = e^{-x} \) in our case.
- Set up the definite integral by subtracting the lower function from the upper one. This gives the height difference between the curves at each point in the interval.
- Compute the integral over the range where the two curves intersect or are confined. In our example, this was from \( x = -2 \) to \( x = 0 \).