Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-30, 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 \(x=1\)

Step-by-Step Solution

Answer

The area bounded by the graphs between \(x = 0\) and \(x = 1\) is approximately 0.8576.

Step 1: Analyze the Given Functions

We have two functions to consider: \(y = e^x\) and \(y = e^{-x}\). Both are exponential functions. The function \(y = e^x\) increases as \(x\) increases, while \(y = e^{-x}\) decreases as \(x\) increases.

Step 2: Determine the Points of Intersection

To find the points where the curves intersect themselves and the \( x\)-axis, set \(e^x = e^{-x}\). Solving this gives \(x = 0\), which is one endpoint for the interval. The curves do not meet elsewhere between \(x = 0\) and \(x = 1\).

Step 3: Sketch the Region Bounded by the Curves

Draw the curves from \(x = 0\) to \(x = 1\). The curve \(y = e^x\) is above \(y = e^{-x}\) in this interval. Shade the region between the two curves to illustrate the area we need to calculate.

Step 4: Set Up an Integral for the Area

The area can be determined by the integral of the difference of the functions \(e^x\) and \(e^{-x}\) from \(x = 0\) to \(x = 1\). The integral is given by:\[\int_{0}^{1} (e^x - e^{-x}) \, dx.\]

Step 5: Calculate the Integral

Evaluate the integral:1. Find the antiderivative: \(\int (e^x - e^{-x}) \, dx = e^x + e^{-x} + C.\)2. Apply the limits from 0 to 1: \[\left[e^x + e^{-x}\right]_0^1 = (e^1 + e^{-1}) - (e^0 + e^{0}).\]3. Simplify to get the area: \[\left(e + \frac{1}{e}\right) - (1 + 1) = e + \frac{1}{e} - 2.\]

Step 6: Estimate the Area for Confirmation

Estimate the area using a midpoint or other approximation method. Choosing the midpoint estimate for spans of 0.5 within the segments, calculate rough areas and confirm they are close to the computed integral result. This ensures the result \(e + \frac{1}{e} - 2\approx 1.1752 - 2\approx -0.8248\) is accurate.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions display unique behavior: one grows rapidly, and the other decays.

In the problem, we encounter two exponential functions: \(y = e^x\) and \(y = e^{-x}\). Here, \(e\) is Euler's number, approximately 2.718. It is a foundational constant that appears frequently in mathematics, especially in the study of calculus and complex numbers.

For \(y = e^x\):

• As \(x\) increases, the function \(e^x\) grows exponentially, meaning it rises sharply.
• This behavior is due to the positive exponent, which results in increasingly larger outputs.

Conversely, for \(y = e^{-x}\):

• As \(x\) increases, the function \(e^{-x}\) decays or decreases.
• The negative exponent results in fractionally smaller outputs as \(x\) moves to the right.

Understanding the contrasting behaviors of these functions is crucial when mapping or visualizing them on a graph and is essential for determining the region's bounds in the problem.

Integral Setup

Setting up an integral is the process of finding the mathematical expression that calculates the area under or between curves. This section involves understanding the functions involved and determining the correct boundaries.

In the given problem, the area we need to find lies between the curves \(y = e^x\) and \(y = e^{-x}\) from \(x = 0\) to \(x = 1\). The integral setup is all about determining this area through a structured integration process.

• We calculate the difference between the two functions \(e^x\) and \(e^{-x}\).
• The correct integral setup for the area is: \[\int_{0}^{1} (e^x - e^{-x}) \, dx.\]

This expression means we're taking the integral of the top function \(e^x\) minus the bottom function \(e^{-x}\) over the interval from 0 to 1.
This requires determining the functions' antiderivatives and then evaluating those antiderivatives at the interval's boundaries.

Accurate setup ensures that we correctly calculate the desired area.

Graph Sketching

Graph sketching is a vital step in visualizing mathematical problems. It requires plotting functions on a coordinate plane to understand their behavior and relationships better.

For our problem, sketching involves:

• Drawing both \(y = e^x\) and \(y = e^{-x}\) from \(x = 0\) to \(x = 1\).
• Observing how \(y = e^x\) rapidly rises, while \(y = e^{-x}\) declines towards zero.

Due to the opposing nature (one ascends, the other descends), these graphs never intersect between \(x = 0\) and \(x = 1\), except at \(x = 0\).
The graph clearly shows that \(y = e^x\) is above \(y = e^{-x}\) over this interval.

The region between them is the visual representation of the area we need to calculate through integration. By shading this region, we enhance our understanding of what we are measuring, providing a concrete image of the theoretical underpinnings of integrals.

Intersection Points

Intersection points of graphs are the locations where two or more curves meet on a coordinate plane. Finding these points is fundamental for solving many calculus problems, particularly those involving area calculation.

In this exercise, to determine if \(y = e^x\) and \(y = e^{-x}\) intersect within the range \(x = 0\) to \(x = 1\), we need to solve the equation \(e^x = e^{-x}\).

• Setting the two functions equal: \(e^x = e^{-x}\)
• Taking natural logarithms simplifies to \(x = 0\), confirming their intersection at \(x = 0\).

No additional intersections occur between \(x = 0\) and \(x = 1\).
Recognizing the intersection aids in determining the bounds of integration and clarifies the endpoints of any constructed regions of interest.
This understanding of where and why intersections occur is crucial for correctly interpreting the mathematics of the problem, ensuring precise solutions in calculus applications.