The question reads:

Evaluate the area of the region bounded by y=sec^2x and y = 4 between x = -pi/3 and x = pi/3.

Thank You!

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- Mar 30th 2009, 04:55 PMMeeklo BracaCould I get some help with finding the area of a region?
The question reads:

Evaluate the area of the region bounded by y=sec^2x and y = 4 between x = -pi/3 and x = pi/3.

Thank You! - Mar 30th 2009, 05:06 PMTheEmptySet
- Mar 30th 2009, 05:07 PMskeeter
- Mar 30th 2009, 05:12 PMMeeklo Braca
I dont know how to integrate that

- Mar 30th 2009, 05:24 PMskeeter
- Mar 30th 2009, 05:31 PMMeeklo Braca
i dont know the anti derivative of sec^2x

- Mar 30th 2009, 05:39 PMskeeter
- Mar 30th 2009, 05:50 PMMeeklo Braca
I gotcha, So I have A = x^2 -tan x, Where do I go from there?

- Mar 30th 2009, 06:07 PMskeeter
- Mar 30th 2009, 06:14 PMMeeklo Braca
I know the fundamental theorm of calculus for non trig functions. The trig throws me for a loop. What form should my answer be in?

- Mar 30th 2009, 06:18 PMTheEmptySet
- Mar 30th 2009, 06:32 PMMeeklo Braca
What form should my final answer be? If I calculate it out its 4.81 but I have a feeling thats not the correct answer.

- Mar 30th 2009, 06:52 PMTheEmptySet

$\displaystyle \int_{-\pi/3}^{\pi/3}4-\sec^2(x)dx=4x-\tan(x) \bigg|_{-\pi/3}^{\pi/3}=4\left( \frac{\pi}{3}\right) -\tan\left( \frac{\pi}{3}\right)-4\left( \frac{-\pi}{3}\right)+\tan\left( \frac{-\pi}{3}\right)= $

$\displaystyle 8\left( \frac{\pi}{3}\right)-2\tan\left( \frac{\pi}{3}\right) \approx 4.91$ - Mar 30th 2009, 07:07 PMMeeklo Braca
OK I will go with that.

Thank you very much for your help and insight!