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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.2.73

73. Find the area between the curves y=ln(x) and y=ln(2x) from x=1 to x=5.

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Identify the two functions given: \( y = \ln(x) \) and \( y = \ln(2x) \). We want to find the area between these curves from \( x = 1 \) to \( x = 5 \).
Determine which function is on top and which is on the bottom in the interval \( [1, 5] \). Since \( \ln(2x) = \ln(2) + \ln(x) \), it is always greater than \( \ln(x) \) for \( x > 0 \). So, \( y = \ln(2x) \) is the upper curve and \( y = \ln(x) \) is the lower curve.
Set up the integral for the area between the curves as \( \int_1^5 [\ln(2x) - \ln(x)] \, dx \).
Simplify the integrand using logarithm properties: \( \ln(2x) - \ln(x) = \ln\left(\frac{2x}{x}\right) = \ln(2) \). So the integral becomes \( \int_1^5 \ln(2) \, dx \).
Evaluate the integral by integrating the constant \( \ln(2) \) over \( [1, 5] \), which is \( \ln(2) \times (5 - 1) \). This gives the area between the curves.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals and Area Between Curves

The area between two curves over an interval is found by integrating the difference of their functions. Specifically, the integral of the upper function minus the lower function from the lower to upper bounds gives the enclosed area.
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