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Chapter 15: Problem 5

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 curve \(y=e^{x}\) and the lines \(y=0, x=0,\) and \(x=\ln 2\)

Short Answer

Expert verified
The area is 1.

Step by step solution

01

Sketch the Region

The curve given is an exponential function, \(y = e^x\), which crosses the y-axis at \(y=1\) and increases rapidly. The lines \(y=0\), \(x=0\), and \(x=\ln 2\) form boundaries for the region. The line \(y=0\) is the x-axis. The line \(x=0\) is the y-axis, and \(x=\ln 2\) is a vertical boundary at \(x=\ln 2\). The region of interest is below \(y = e^x\), above \(y = 0\), right of \(x = 0\), and left of \(x = \ln 2\).
02

Set Up the Double Integral

The region is vertically simple, meaning the y-boundaries are functions of x. Therefore, set up the double integral using x-integrals as outer integrals.The iterated integral is:\[\int_{x=0}^{x=\ln 2} \left( \int_{y=0}^{y=e^x} dy \right) dx\]
03

Evaluate the Inner Integral

First, evaluate the inner integral:\[\int_{y=0}^{y=e^x} dy = [y]_{0}^{e^x} = e^x - 0 = e^x\]
04

Evaluate the Outer Integral

Substitute the result from the inner integral into the outer integral:\[\int_{x=0}^{x=\ln 2} e^x \, dx\]Evaluate the integral:\[[e^x]_{0}^{\ln 2} = e^{\ln 2} - e^0 = 2 - 1 = 1\]
05

Conclusion

The area of the region bounded by the curve and lines is \(1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. In our problem, the curve is represented by the exponential function \( y = e^x \). This curve illustrates a number of characteristics that make exponential functions distinctive:
  • Exponential Growth: As \( x \) increases, \( e^x \) rapidly increases because the base \( e \) (approximately 2.718) is greater than 1.
  • Intercept: At \( x = 0 \), \( e^0 = 1 \), meaning the curve crosses the y-axis at 1.
These features make exponential functions crucial in modeling growth processes and decay phenomena in nature and finance. In the given exercise, the exponential curve is bounded above the line \( y = 0 \) and vertical lines \( x = 0 \) and \( x = \ln 2 \), forming a region of interest. Understanding how \( e^x \) behaves helps us sketch and identify these bounded regions correctly.
Definite Integrals
Definite integrals allow us to calculate the exact area under a curve over a specific interval. In this exercise, we set up a double integral to find the area of the region bounded by the curve \( y = e^x \) and lines \( y = 0 \), \( x = 0 \), and \( x = \ln 2 \). The steps involved are key to understanding how integration works:

  • Inner Integral: This accounts for the area in the vertical direction, from \( y = 0 \) to \( y = e^x \), for each 'slice' or cross-section of the region.
  • Outer Integral: Summed over the x-interval from \( x = 0 \) to \( x = \ln 2 \), this gathers all the slices together to compute the total area.
Evaluating the integral gives us \( e^{\ln 2} - e^0 = 2 - 1 = 1 \). So, the area of the region is 1. With definite integrals, the bounds are important, as they define the limits over which we calculate the area. This mathematical process captures the region as a whole by integrating the smaller, simpler, vertical sections.
Area Between Curves
Finding the area between curves is a common application of calculus, often involving double integrals. In our specific problem, we are interested in the area between the curve \( y = e^x \) and the x-axis \( y = 0 \), bounded by two vertical lines \( x = 0 \) and \( x = \ln 2 \).

This area is calculated as follows:
  • The curve and line form a closed region, making it simple to apply double integration to find the enclosed area.
  • Visually, you can view this space between the curve and x-axis as a collection of thin vertical strips, each integrated and combined to form the whole area.
  • In this problem, because \( y = 0 \) forms a constant lower boundary interacting with the exponential curve, the integral simplifies neatly with the bounds for x and y clearly defined.
By using integrals, particularly double integrals for this multi-dimensional space, we manage to break down the calculation into simpler parts. This method is valuable when working with more complex functions and graphs.

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