Problem 72 - Thomas Calculus

Question

Find the area of the "triangular" region in the first quadrant that is bounded above by the curve \(y=e^{x / 2},\) below by the curve \(y=e^{-x / 2},\) and on the right by the line \(x=2 \ln 2 .\)

Step-by-Step Solution

Answer

The area of the triangular region is 3 square units.

Step 1: Understand the Boundaries

The problem describes a region in the first quadrant bounded by three curves: above by \(y=e^{x/2}\), below by \(y=e^{-x/2}\), and on the right by the vertical line \(x=2\ln 2\). First, it's crucial to visualize this setup on the xy-plane.

Step 2: Identify the Intersection Points

To find the area, identify where the curves \(y=e^{x/2}\) and \(y=e^{-x/2}\) intersect. Set the equations equal: \[e^{x/2}=e^{-x/2}\]. This simplifies to \[e^{x}=1\], yielding x=0 as the intersection point. At \(x=0\), both curves equal 1, hence the point \((0,1)\).

Step 3: Determine Functional Relationships

We are now left with the region of integration from \(x=0\) to \(x=2\ln 2\). For these x values, calculate differences in y values for both functions: \(y_1 = e^{x/2}\) and \(y_2 = e^{-x/2}\).

Step 4: Set up the Integral

The area of the region is determined by the integral of the difference of the functions. Express this area integral as follows: \[ A = \int_{0}^{2\ln 2} (e^{x/2} - e^{-x/2}) \, dx \]

Step 5: Solve the Integral

Solve the integral: \[ A = \int_{0}^{2\ln 2} e^{x/2} \, dx - \int_{0}^{2\ln 2} e^{-x/2} \, dx \]The antiderivatives are: \( \int e^{x/2}dx = 2e^{x/2} \) and \( \int e^{-x/2}dx = -2e^{-x/2} \).

Step 6: Substitute Limits and Evaluate

Substitute the upper and lower limits into the antiderivatives:\[ 2e^{x/2}\Big|_{0}^{2\ln 2} = 2e^{(2\ln 2)/2} - 2e^{0} = 4 - 2 = 2 \]\[ -2e^{-x/2}\Big|_{0}^{2\ln 2} = -2e^{-(2\ln 2)/2} + 2e^{0} = -1 + 2 = 1 \]Therefore, the area is \[2 + 1 = 3\].

Key Concepts

Integration

Integration is a fundamental concept in calculus. It involves finding the whole from knowing the rate of change. Think about it like adding up an infinite number of infinitesimally small quantities.
In the context of our exercise, integration allows us to find the area between two curves. To achieve this:

• We set up an integral to calculate the area under a curve or between curves.
• The integral bounds determine where the calculation starts and ends on the x-axis.
• We find antiderivatives to help solve the integral. These are functions whose derivatives give us the original functions.

Understanding integration is key to solving many problems in calculus, especially those involving areas and accumulated quantities.

Area under a curve

The area under a curve is a specific application of integration. By evaluating an integral, we determine how much space is covered between a curve and the x-axis, or between two curves.
In our exercise, we calculated the area between the exponential curves and the vertical line. Here’s how it was done:

• We identify the functions that form our boundaries, top and bottom.
• We calculate the difference between these functions over a specific interval.
• Using integration, we find the cumulative area between the curves from one x-value to another.

This process involves seeing integrals as accumulation of infinitesimally small rectangles spanning from the lower curve to the upper curve, which summed up, give the desired total area.

Exponential functions

Exponential functions are mathematical expressions involving constants raised to a variable power. They are usually written in the form \(y = a^{x}\), where \(a\) is a positive constant.
In our exercise, we worked with \(e^{x/2}\) and \(e^{-x/2}\), which are two exponential functions:

• These functions show exponential growth and decay based on the sign and magnitude of the exponent.
• Exponential functions tend to rise indefinitely or decrease toward zero.
• They have unique derivatives and antiderivatives, making them straightforward to integrate and differentiate.

Understanding their properties is crucial in determining the behavior of functions and solving problems associated with growth, decay, and accumulation in calculus.