Region between two curves, trig twist?

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In summary, when finding the area between the curves y = tan 3x and y = 2 sin 3x in the interval −π/9 ≤ x ≤ π/9, the correct approach is to split the integral into two parts and add the magnitudes of the values. The answer (2/3)*(ln(2)-1) was close, but had a sign issue in the parentheses. The integral of the left/bottom region is 1/3(1-ln(2)), which is positive.
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stripedcat
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 

FAQ: Region between two curves, trig twist?

What is the "region between two curves, trig twist"?

The region between two curves, trig twist, refers to the area enclosed between two curves on a graph, where one of the curves is a trigonometric function such as sine or cosine.

How do you find the area of the region between two curves with a trig twist?

To find the area of this region, you will need to use integration techniques such as the definite integral. First, identify the points of intersection between the two curves. Then, set up the integral with the appropriate limits of integration and integrate the difference between the two curves.

What is the significance of the trig twist in this region?

The trig twist in this region can affect the shape and size of the enclosed area. It is important to carefully consider the behavior of the trigonometric function and how it intersects with the other curve in order to correctly calculate the area.

Can the region between two curves with a trig twist have a negative area?

Yes, it is possible for the region between two curves with a trig twist to have a negative area if the curves intersect in a way that creates a region below the x-axis. In this case, the area would be represented by a negative value in the integral.

Are there any special cases to consider when finding the area of the region between two curves with a trig twist?

Yes, there are some special cases to consider, such as when the curves intersect at multiple points or when the trigonometric function has multiple periods within the given interval. It is important to carefully analyze the graph and adjust the limits of integration accordingly in these cases.

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