Problem 60 - Calculus and its applications

Question

Find the area of the region bounded by the given graphs. $$ y=e^{x}, y=e^{-x}, x=-2 $$

Step-by-Step Solution

Answer

The area is \( 2 - (e^2 + e^{-2}) \).

Step 1: 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.

Step 2: 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 \( x = 0 \).

Step 3: 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.

Step 4: 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 \( e^x + e^{-x} \).

Step 5: 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: \( (2) - (e^2 + e^{-2}) \).

Key Concepts

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:

• 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.

Remember, the result of a definite integral is a number that represents the total area between your defined curves and within your given limits.

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:

• 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.

When dealing with problems involving exponential functions, it's essential to understand their behavior and interaction, especially in areas like integration and finding intersection points.

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:

• 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 \( x = 0 \).

The result tells you the total enclosed 'area' between these curves. This method is crucial not only for academic problems but also in practical applications such as physics or engineering, where modeling and interpreting data regions play a critical role.