Problem 38
Question
For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection points analytically, use a calculator to approximate the intersection points with three decimal places and determine the approximate area of the region. $$ [T]x=e^{y} \text { and } y=x-2 $$
Step-by-Step Solution
Verified Answer
The approximate area is 3.659 square units.
1Step 1: Understand the Problem
We are asked to find the area enclosed between the curves given by the equations \(x = e^y\) and \(y = x - 2\). This requires us to find the points of intersection and then set up an appropriate integral to calculate the area between the curves.
2Step 2: Find the Intersection Points Analytically
To find the intersection points, set the two equations equal to each other. Substitute \(y = x - 2\) into the equation \(x = e^y\):\[x = e^{x-2}.\] This equation cannot be solved analytically for \(x\), thus we must use a calculator to find the approximate intersection points.
3Step 3: Approximate Intersection Points Using Calculator
Use a calculator to solve the equation \(x = e^{x-2}\). Approximating, we find the intersection points to be around \((2.134, 0.134)\) and \((5.146, 3.146)\).
4Step 4: Set Up the Integral for Area Between Curves
The area can be found by integrating along the \(x\)-axis from the \(x\)-coordinate of the first intersection point to the \(x\)-coordinate of the second. Use the integral formula: \[\text{Area} = \int_{x_1}^{x_2} (\text{top function} - \text{bottom function}) \, dx.\] Here, the top function is \(x\) (since \(x = e^y\) is implicitly solved as \(y = \ln(x)\)) and the bottom is \(y = x - 2\).
5Step 5: Evaluate the Integral
The integral for the area will be: \[\int_{2.134}^{5.146} \ln(x) - (x - 2) \, dx.\]Calculating this integral either numerically using a calculator gives us an approximate area.
6Step 6: Calculate the Area Using Numerical Methods
Using numerical methods on a calculator, the integral evaluates to approximately 3.659. Thus, the approximate area of the region bounded by these curves is around 3.659 square units.
Key Concepts
Area between curvesIntersection pointsNumerical methodsIntegral evaluation
Area between curves
To find the area between two curves, we need to understand the concept of integration between two boundaries. The curves can be defined by equations such as functions. When we have two curves, the area between them is the region enclosed by their paths. Often, one function will be on top and the other at the bottom across a specific interval.
It's important to choose the correct boundaries to ensure we are considering the right segment of area. Integration involves taking small slices or "strips" of this area and summing them up. The more precise the calculation, the more accurate the area is. In many mathematical problems, especially when calculating areas between curves, this process helps us understand the space enclosed by different paths on a graph.
With the given equations, once we determine the intersection points, we can define our interval for integration and solve for the area.
It's important to choose the correct boundaries to ensure we are considering the right segment of area. Integration involves taking small slices or "strips" of this area and summing them up. The more precise the calculation, the more accurate the area is. In many mathematical problems, especially when calculating areas between curves, this process helps us understand the space enclosed by different paths on a graph.
With the given equations, once we determine the intersection points, we can define our interval for integration and solve for the area.
Intersection points
Finding intersection points is crucial because they determine the boundaries of the area we need to calculate. Intersection points are where the graphs of the functions cross each other. They can be found by setting the equations equal and solving for the variables.
Unfortunately, some equations cannot be easily solved analytically. This is when calculators or numerical algorithms become valuable tools. You can use calculators to approximate intersection points to a few decimal places, which, although not exact, gives a very close approximation.
Once the intersection points are known, you can use these as limits in your integral for finding the area between curves.
Unfortunately, some equations cannot be easily solved analytically. This is when calculators or numerical algorithms become valuable tools. You can use calculators to approximate intersection points to a few decimal places, which, although not exact, gives a very close approximation.
Once the intersection points are known, you can use these as limits in your integral for finding the area between curves.
Numerical methods
Numerical methods are techniques used to approximate solutions where exact solutions are difficult to find. For finding the area between curves, numerical integration can be used when the antiderivative is hard or impossible to find analytically.
Techniques such as the trapezoidal rule or Simpson's rule divide the integral into small segments and estimate their areas. Calculators or computer software can quickly perform these calculations for more accuracy.
In our problem, numerical methods allow us to handle complex functions and provide meaningful results without deep analytical derivation.
Techniques such as the trapezoidal rule or Simpson's rule divide the integral into small segments and estimate their areas. Calculators or computer software can quickly perform these calculations for more accuracy.
- Trapezoidal Rule: Uses linear interpolation to approximate the region under the curve.
- Simpson's Rule: Utilizes polynomial functions for a better approximation, usually resulting in more accurate results.
In our problem, numerical methods allow us to handle complex functions and provide meaningful results without deep analytical derivation.
Integral evaluation
Integral evaluation is the process of calculating an integral to find the area under a curve. For area between curves, the integral is set up to subtract the equation of the bottom curve from the equation of the top curve over a specified interval.
After setting up the integral properly with upper and lower limits based on the intersection points, you can use this setup to evaluate either analytically or numerically. In our case where analytical means are tough, numerical solutions will provide the region's approximate area.
The core idea here is: the integral simplifies the problem into a form where you can manage the functions involved, even when exact algebraic solutions are hard to achieve. Ultimately, integral evaluation with technological help provides an avenue for solving otherwise intractable mathematical problems.
After setting up the integral properly with upper and lower limits based on the intersection points, you can use this setup to evaluate either analytically or numerically. In our case where analytical means are tough, numerical solutions will provide the region's approximate area.
The core idea here is: the integral simplifies the problem into a form where you can manage the functions involved, even when exact algebraic solutions are hard to achieve. Ultimately, integral evaluation with technological help provides an avenue for solving otherwise intractable mathematical problems.
Other exercises in this chapter
Problem 36
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
View solution Problem 37
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
View solution Problem 39
For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection point
View solution Problem 40
For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection point
View solution