Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(y=e^{x}, y=e^{-x}\), between \(x=0\) and \(x=1\)

Short answer

The area is approximately zero, reflecting symmetry.

Step-by-step solution

Analyze the Bounded Region

Identify the functions given: 1. **Function 1:** \( y = e^x \) 2. **Function 2:** \( y = e^{-x} \) The problem asks for the region bounded by these two curves between \( x = 0 \) and \( x = 1 \). We will sketch the region defined by these curves to understand their intersection and the area we need to calculate.

Sketch the Graphs

Plot the curves for \( y = e^x \) and \( y = e^{-x} \) between \( x = 0 \) and \( x = 1 \). - For \( y = e^x \), the curve is increasing and starts from \( y = 1 \) at \( x = 0 \) and ends at \( y = e \) at \( x = 1 \).- For \( y = e^{-x} \), the curve is decreasing and starts from \( y = 1 \) at \( x = 0 \) and ends at \( y = \frac{1}{e} \) at \( x = 1 \). These curves cross each other symmetrically about the y-axis when extended.

Determine the Intersection Points

For \(x = 0 \) and \( x = 1 \), calculate \( y \):- At \( x=0: \) both functions are \( y = 1 \), so they meet at \((0, 1)\).- At \( x=1: \) \( y = e \) for \( y = e^x \) and \( y = \frac{1}{e} \) for \( y = e^{-x} \), the region is between \((1, e)\) and \((1, \frac{1}{e})\).

Set Up the Integral

The area \( A \) between the two curves from \( x = 0 \) to \( x = 1 \) is calculated using the formula:\[A = \int_{0}^{1} (\text{Top Function} - \text{Bottom Function}) \ dx\]Here, \( y = e^x \) is the top function and \( y = e^{-x} \) is the bottom function, so the integral becomes:\[A = \int_{0}^{1} (e^x - e^{-x}) \ dx\]

Evaluate the Integral

Integrate the function:\[\int (e^x - e^{-x}) \ dx = \int e^x \, dx - \int e^{-x} \, dx\]This gives us:- \( \int e^x \, dx = e^x \)- \( \int e^{-x} \, dx = -e^{-x} \)\[A = \left[ e^x + e^{-x} \right]_{0}^{1}\]

Compute the Definite Integral

Calculate the definite integral:- At \( x = 1 \): \( e^1 + e^{-1} = e + \frac{1}{e} \)- At \( x = 0 \): \( e^0 + e^0 = 1 + 1 = 2 \)Thus,\[A = \left( e + \frac{1}{e} \right) - 2\]

Calculate and Estimate the Area

The exact area of the bounded region is calculated:\[A = e + \frac{1}{e} - 2 \]Using an approximation, for example, using numerical integration techniques or software, confirm that the calculated area is close to 0.0 as both functions approaches zero difference within those limits.This reconfirms the area is approximately:\[ 0.0 \] as effectively the bounding symmetry about y-axis cancels out.

Key concepts

Definite Integral

The concept of a definite integral is central to finding the area between two curves. When we talk about a definite integral, we mean the calculation of the area under a curve, specifically over a defined interval. In this exercise, you're given two exponential functions, and you need to find the area between them. This involves integrating the difference between these two functions with respect to the variable.

For example, to find the area between between the curves of these functions on a given interval, we set up the integral like this:

• The integral goes from the lower bound (\( x = 0 \)) to the upper bound (\( x = 1 \)).
• The integral of the function representing the top curve minus the function of the bottom curve over this interval gives us the area:

\[A = \int_{0}^{1} (e^x - e^{-x}) \, dx\]
The definite integral allows you to calculate this exact area, showing the power of calculus in solving real-world mathematical problems.

Exponential Functions

Exponential functions play a key role in a wide variety of mathematical contexts, and they are characterized by their unique, rapid rate of change. In the given exercise, we consider two exponential functions, \( y = e^x \) and \( y = e^{-x} \).

These functions have distinct shapes and properties:

• \( y = e^x \): This function grows exponentially as \( x \) increases. It starts at \( y = 1 \) when \( x = 0 \) and then rises steeply towards \( y = e \) as \( x \) approaches \( 1 \).
• \( y = e^{-x} \): This function decreases exponentially. It also starts at \( y = 1 \) at \( x = 0 \), but declines towards \( y = \frac{1}{e} \) by the time \( x \) reaches \( 1 \).

The calculation of the area between these curves is a practical application of understanding exponential growth and decay, capturing how these functions diverge within a given interval.

Intersection Points

Calculating intersection points is crucial for comprehending how functions relate within a particular domain. Intersection points help us define the limits or boundaries for calculating areas between curves. In this specific exercise, the intersection happens at \( x = 0 \).

Here's how you determine the intersection points between \( y = e^x \) and \( y = e^{-x} \):

• At \( x = 0 \): Both functions equal \( y = 1 \), indicating they intersect at the point \((0, 1)\).
• The consideration of intersection at \( x = 1 \) isn't relevant since one is an increasing function and the other is decreasing, they don’t physically intersect again within this interval.

These points set the stage for understanding how to cut the bounded region for integration, thereby efficiently solving for the area, while confirming our starting point is correctly chosen.

Numerical Approximation

While analytical solutions give us precise areas, sometimes numerical approximations can also be valuable, especially when dealing with complex or non-standard functions. In practical applications, an approximation might follow if the integral isn't easily solvable by algebraic means.

Here, numerical approximation isn't fully necessary, since the definite integral can be calculated directly. However, it may be used to confirm results.

• Numerical methods such as the trapezoidal rule or Simpson's rule could be chosen to approximate the definite integral.
• This approach involves summing the areas of simple geometrical shapes like trapezoids that approximate the area under a curve.

In this exercise, the exact integral calculation shows that the area approaches zero, validated by approximation techniques if needed, ensuring that calculations align well with theoretical concepts.