Total area enclosed between ln(x) and sin^2(2x)-cos(3x)+1

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Discussion Overview

The discussion revolves around finding the total area enclosed between the functions ln(x) and sin²(2x) - cos(3x) + 1. Participants explore the methods for determining the points of intersection and setting up definite integrals to calculate the area, while also expressing concerns about the complexity of the problem.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes the process of solving ln(x) = sin²(2x) - cos(3x) + 1 to find intersection points and setting up definite integrals, expressing concern about needing to add 10 integrals for the total area.
  • Another participant questions the choice of functions due to the lack of an analytic solution and the frequent intersections, suggesting that it leads to complicated numeric integration.
  • A later reply reiterates the concern about the problem's origin and the unusual number of intersections, noting that it was derived from an old worksheet.
  • Some participants mention the necessity of managing signs in the integration process, indicating that this adds to the complexity of finding a solution.

Areas of Agreement / Disagreement

Participants generally express uncertainty about the efficiency of the proposed methods for solving the problem, with no consensus on a faster approach or resolution to the complexity involved.

Contextual Notes

The discussion highlights the challenges of numeric integration in cases where analytic solutions are not available, as well as the implications of having multiple intersections on the setup of integrals.

Saracen Rue
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I understand the theory behind this type of question well enough; you solve ln(x)=sin^2(2x)-cos(3x)+1 to find the x values at the points of intersection, and then set up definite integrals over the domains of said x-values, subtracting whichever function is below the other for a specific domain.

However, when doing this particular question I realized I needed to add 10 definite integrals together to obtain the total area, which seems rather excessive to me. So I was just wondering if there's a faster way of doing this question? Thank you for your help. (And in case you're wondering the answer should be 9.7435)
 
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Where does that problem come from? The equation doesn't even have an analytic solution, so it has to be ugly numeric integration, and I don't see the point of choosing functions that intersect so often.
 
mfb said:
Where does that problem come from? The equation doesn't even have an analytic solution, so it has to be ugly numeric integration, and I don't see the point of choosing functions that intersect so often.

It was off an old work sheet from a couple of decades back that I got given as revision because I finished everything else early. I don't have the sheet anymore but I had this question written down because I found it particularly odd. The graphs intersect 11 times which is more than usual but still not too bad. I'm just curious though if there's a faster way to get the answer than setting up 10 definite integrals
 
Nothing that would really help, as you have to take care of the signs in some way.
 
mfb said:
Nothing that would really help, as you have to take care of the signs in some way.
Okay, thank you for the help
 

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