Problem 10 - Thomas Calculus

Question

In Exercises \(1 - 12 ,\) sketch the region bounded by the given lines and curves. Then express the region's area as an iterated double integral and evaluate the integral. The lines \(y = 1 - x\) and \(y = 2\) and the curve \(y = e ^ { x }\)

Step-by-Step Solution

Answer

The area is \( 2\ln(2) + \frac{1}{e} \).

Step 1: Identify the region bounded by the given equations

The given equations are the lines \( y = 1 - x \), \( y = 2 \), and the curve \( y = e^x \). To find the region bounded by these graphs, we need to determine where these lines and curve intersect. Calculate these intersection points by equating the functions two at a time.

Step 2: Find Points of Intersection

First, find the intersection of \( y = 1 - x \) and \( y = e^x \):\[ 1 - x = e^x \]Solving this equation numerically, we find that they do not intersect within our visible range. Next, find the intersection of \( y = e^x \) and \( y = 2 \):\[ e^x = 2 \]This occurs at \( x = \ln(2) \). Lastly, find the intersection of \( y = 1 - x \) and \( y = 2 \):\[ 1 - x = 2 \]This solves to \( x = -1 \). Thus, the bounded region involves these points and the lines/curve as described.

Step 3: Set up the iterated double integral for the area

The region is bounded by \( x = -1 \) and \( x = \ln(2) \). The upper boundary is \( y = 2 \) and it transitions from \( y = e^x \) to \( y = 1 - x \) based on the \( x \) value. Set up the double integral for the bounded region:\[ \int_{-1}^{\ln(2)} \int_{e^x}^{2} dy \, dx \].

Step 4: Evaluate the inner integral

Evaluate the inner integral with respect to \( y \):\[ \int_{e^x}^{2} dy = [y]_{e^x}^{2} = (2 - e^x) \].The result simplifies the iterated double integral to:\[ \int_{-1}^{\ln(2)} (2 - e^x) \, dx \].

Step 5: Evaluate the outer integral

Now, integrate \( 2 - e^x \) with respect to \( x \):\[ \int_{-1}^{\ln(2)} (2 - e^x) \, dx = \int_{-1}^{\ln(2)} 2 \, dx - \int_{-1}^{\ln(2)} e^x \, dx \].First, calculate \( \int_{-1}^{\ln(2)} 2 \, dx = 2[x]_{-1}^{\ln(2)} = 2(\ln(2) + 1) = 2\ln(2) + 2 \).Second, calculate \( \int_{-1}^{\ln(2)} e^x \, dx = [e^x]_{-1}^{\ln(2)} = (2 - \frac{1}{e}) \).So, the iterated integral results in:\[ 2\ln(2) + 2 - 2 + \frac{1}{e} = 2\ln(2) + \frac{1}{e} \].

Key Concepts

Bounded Regions

When considering double integrals over an area, one crucial step is defining the region where the integration will take place. This is known as the bounded region. Here, the bounded region involves the lines and curves described by the equations: \( y = 1 - x \), \( y = 2 \), and \( y = e^x \).
These boundaries help define the specific portion of the plane that we are interested in. By understanding which area is enclosed by these functions, we can proceed with applying double integrals effectively. The process involves sketching the area to visualize intersections and transitions between boundaries.
Defining these limits clearly is important to solve the integral properly, and it helps ensure that the integration covers exactly the intended area.

Intersection Points

Identifying intersection points is a key task when working with bounded regions. These points determine where the boundaries meet and how they interact with one another.
To find intersection points, we equate the given functions pairwise. For instance:

• To find where \( y = 1 - x \) and \( y = e^x \) intersect, solve \( 1 - x = e^x \). In this case, a graphical or numerical solution shows no intersection within the visible range.
• For \( y = e^x \) and \( y = 2 \), solve \( e^x = 2 \), resulting in \( x = \ln(2) \).
• To find where \( y = 1 - x \) and \( y = 2 \) intersect, solve \( 1 - x = 2 \), giving \( x = -1 \).

These intersection points are essential to define the integration limits for calculating the area.

Exponential Functions

Exponential functions, like \( y = e^x \), are pivotal in many integration problems. Here, it forms part of the boundary for the region we are interested in. Exponential functions have distinctive properties:

• They always pass through the point \((0, 1)\).
• They increase rapidly as \( x \) becomes large.
• They never touch the x-axis, only approaching it as \( x \) becomes negative infinity.

The behavior of exponential functions affects how they intersect with other lines or curves, influencing the overall shape of the region under consideration. Understanding these properties helps when setting up the limits for the double integral.

Integral Evaluation

Once the bounded region and intersection points are defined, the next step is to set up and evaluate the associated double integral. A double integral calculates the volume under the surface over the specified region.
In our setup, the integral is \( \int_{-1}^{\ln(2)} \int_{e^x}^{2} dy \, dx \), where we integrate "inside-out". This means we first evaluate the inner integral:

• Calculate \( \int_{e^x}^{2} dy = 2 - e^x \).

The result of this integration simplifies the problem to an easier form:
\( \int_{-1}^{\ln(2)} (2 - e^x) \, dx \). From there, the outer integral is computed:

• Break it into two separate integrals: \( \int 2 \, dx \) and \( - \int e^x \, dx \).

This step-by-step breakdown is crucial for finding the correct solution, highlighting the analytical skills used in calculus. Completing this evaluation gives the necessary result for the problem.