Region between two curves, trig twist?

[stripedcat]

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?

 

[Jameson]

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.

 

[Prove It]

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.

 

[stripedcat]

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?

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.

 

[Jameson]

How is that negative? 1/3(1-ln(2))>0.

http://www.wolframalpha.com/input/?i=1/3(1-ln(2))