Problem 12 - Calculus

Question

Find where the curves in \(1-12\) intersect, draw rough graphs, and compute the area between them. $$ y=e \text { and } y=e^{x} \text { and } y=e^{-x} $$

Step-by-Step Solution

Answer

The area between the curves is \(2e - 2\).

Step 1: Identify the Intersection Points

The curves given are \(y = e\), \(y = e^x\), and \(y = e^{-x}\). To find their intersection points, set each function equal to each of the others one by one. \[e^x = e\] implies \[x = 1\]. Similarly, \[e^{-x} = e\] implies \[-x = 1\] or \[x = -1\]. Therefore, the curves intersect at \(x = -1\) and \(x = 1\).

Step 2: Draw Rough Graphs

Sketch the graph of \(y = e\), which is a horizontal line at \(y = e\). The graphs of \(y = e^x\) and \(y = e^{-x}\) are exponential functions that intersect \(y = e\) at \(x = 1\) and \(x = -1\), respectively. \(y = e^x\) increases as \(x\) increases from 0, and \(y = e^{-x}\) decreases as \(x\) increases from 0.

Step 3: Set Up Integral for Area

To find the area between the curves from \(x = -1\) to \(x = 1\), we compute the integrals of the difference between the functions. Over this interval, the area between \(y = e\) and \(y = e^x\) or \(y = e^{-x}\) is the integral of \(e - e^x\) from \(-1\) to \(0\) and \(e - e^{-x}\) from \(0\) to \(1\).

Step 4: Calculate the Integrals

Compute the integral: \(\int_{-1}^{0}(e - e^x) \, dx + \int_{0}^{1}(e - e^{-x}) \, dx\). This gives: \[= \left[e x - e^x \right]_{-1}^{0} + \left[e x + e^{-x} \right]_{0}^{1}\] This simplifies to \([e(0) - e^0 - e(-1) + e^{-(-1)}] + [e(1) - e^{-1} + e^{-1} - e(0)]\).

Step 5: Simplify the Evaluations

Substitute the values: \([0 - 1 + \frac{e}{e} + e] + [e - \frac{1}{e} + \frac{1}{e} - e]\) which simplifies further to \([-1 + 1 + e] + [e - 1]\).

Step 6: Final Calculation

Simplify the expression: \((e - 1) + (e - 1) = 2e - 2\). Therefore, the area between the curves is \(2(e - 1) = 2e - 2\).

Key Concepts

Intersection Points

Understanding where curves intersect is crucial when analyzing their relationships. Here, we examine the curves given: \(y = e\), \(y = e^x\), and \(y = e^{-x}\). The intersection points of these curves are points where their equations equal one another. To find these:

• Set \(e^x = e\), giving \(x = 1\).
• Set \(e^{-x} = e\), resulting in \(-x = 1\) or \(x = -1\).

This means our curves intersect at \(x = -1\) and \(x = 1\). Finding these intersection points is vital since they define the limits for calculating areas between curves, which is the primary goal of this exercise.

Exponential Functions

Exponential functions exhibit unique characteristics compared to linear or polynomial functions. The functions \(y = e^x\) and \(y = e^{-x}\) are both exponential:

• \(y = e^x\) starts increasing rapidly from left to right, representing exponential growth.
• \(y = e^{-x}\) portrays exponential decay, decreasing as \(x\) increases.

These increasing and decreasing behaviors make exponential functions excellent for modeling real-world phenomena, like growth and decay. Their distinct shapes also determine how they interact with other functions, such as the horizontal line \(y = e\), and help delineate regions enclosed by these curves.

Definite Integral

Integrals can be used to calculate the area under a curve between two points. A definite integral specifies both the function and the interval over which to integrate. Here, the definite integral helps us compute area between curves. We set up our integrals to find the area between:

• \(y = e\) and \(y = e^x\) from \(x = -1\) to \(0\)
• \(y = e\) and \(y = e^{-x}\) from \(x = 0\) to \(1\)

Mathematically, this is written as:\[\int_{-1}^{0}(e - e^x) \, dx + \int_{0}^{1}(e - e^{-x}) \, dx\]The integral captures the collective area under the influence of both exponential curves, validating the result when both calculations are combined.

Area Between Curves

The computation of the area between curves involves subtracting one function from another, then integrating over a specified interval. This measure highlights how much space lies between two curves along the \(x\)-axis. For our curves \(y = e\), \(y = e^x\), and \(y = e^{-x}\), we:

• Take \(e - e^x\) from \(x = -1\) to \(0\) to capture the slice between \(y = e^x\) and the horizontal line.
• Calculate \(e - e^{-x}\) from \(x = 0\) to \(1\) for the opposite curve.

Initially, it seems we have multiple regions, but combining these, the result reflects the complete area sandwiched between the overall curves. The final result, \(2(e - 1) = 2e - 2\), represents the totality of this space indicating effective integration.