Region between two curves, trig twist?

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Discussion Overview

The discussion revolves around finding the area between the curves defined by the equations y = tan(3x) and y = 2 sin(3x) over the interval from -π/9 to π/9. Participants explore the correct method to calculate this area, considering the implications of signed areas in definite integrals.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant expresses confusion about their calculated area, which was (2/3)*(ln(2)-1), and seeks clarification on their mistake.
  • Another participant suggests that there is a sign issue in the integral calculation and proposes that the integral of the left region yields 1/3(1-ln(2)), indicating a need for a sign change when doubling the area.
  • A later reply emphasizes that the integral values on either side of zero will cancel out if computed as a single integral, leading to a total of zero, and suggests splitting the integral into two parts to find the magnitudes of the areas instead.
  • One participant questions the assertion that the area could be negative, arguing that 1/3(1-ln(2)) is positive based on their calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating the area. There are competing views regarding the treatment of signed areas in integrals and whether the area can be negative.

Contextual Notes

Participants express uncertainty about the implications of negative areas in the context of definite integrals and the necessity of considering magnitudes when calculating total area.

stripedcat
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View attachment 2694

y = tan 3x, y = 2 sin 3x, −π/9 ≤ x ≤ π/9

It's looking for the area in blue there. Now obviously the -pi/9 to 0 is one region, then 0 to pi/9 is the other, but presumably you could find the area of one then double it (which is what I tried to do) but it wasn't right.

The incorrect answer I had was (2/3)*(ln(2)-1).

Could somebody tell me what I'm doing wrong?
 

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Plugging that integral into Wolfram Alpha it looks like you are very close, but have a sign issue in the parentheses. The integral of the left/bottom region is 1/3(1-ln(2)). Doubling this is close to your answer, but with a sign change.
 
stripedcat said:
https://www.physicsforums.com/attachments/2694

y = tan 3x, y = 2 sin 3x, −π/9 ≤ x ≤ π/9

It's looking for the area in blue there. Now obviously the -pi/9 to 0 is one region, then 0 to pi/9 is the other, but presumably you could find the area of one then double it (which is what I tried to do) but it wasn't right.

The incorrect answer I had was (2/3)*(ln(2)-1).

Could somebody tell me what I'm doing wrong?

The sign of the integral value to the left of 0 will be negative, and the sign of the integral value to the right of 0 will be positive. So if you do the entire integral at once, the two so-called "signed areas" will cancel out and leave you a value of 0 for the integral.

If you actually want the area, then you need to split it up into two integrals and then add the MAGNITUDES of the values you have found.
 
Jameson said:
Plugging that integral into Wolfram Alpha it looks like you are very close, but have a sign issue in the parentheses. The integral of the left/bottom region is 1/3(1-ln(2)). Doubling this is close to your answer, but with a sign change.

Yeah, it becomes negative, but a negative area?

Prove It said:
The sign of the integral value to the left of 0 will be negative, and the sign of the integral value to the right of 0 will be positive. So if you do the entire integral at once, the two so-called "signed areas" will cancel out and leave you a value of 0 for the integral.

If you actually want the area, then you need to split it up into two integrals and then add the MAGNITUDES of the values you have found.

This is what I was attempting to do, it didn't work. I don't know what I did wrong.
 

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