Math Help - Could I get some help with finding the area of a region?

1. Could I get some help with finding the area of a region?

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

Thank You!

2. Originally Posted by Meeklo Braca

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

Thank You!
sketch the graph and you will see that y=4 is the upper curve so we get..

$\int _{-pi /3}^{\pi /3}(4-\sec^2{x})dx$

Just integrate from here

3. Originally Posted by Meeklo Braca

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

Thank You!
$A = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 4 - \sec^2{x} \, dx
$

integrate and evaluate.

4. I dont know how to integrate that

5. Originally Posted by Meeklo Braca
I dont know how to integrate that
what exactly is it that you don't know?

don't know the antiderivatives of $4$ and $\sec^2{x}$ ?

don't know how to use the fundamental theorem of calculus?

6. i dont know the anti derivative of sec^2x

7. Originally Posted by Meeklo Braca
i dont know the anti derivative of sec^2x
what is the derivative of $\tan{x}$ ?

8. I gotcha, So I have A = x^2 -tan x, Where do I go from there?

9. Originally Posted by Meeklo Braca
I gotcha, So I have A = x^2 -tan x, Where do I go from there?
that is incorrect ... the antiderivative of $4 - \sec^2{x}$ is $4x - \tan{x}$.

last step is to evaluate the definite integral using the fundamental theorem of calculus.

10. 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?

11. Originally Posted by Meeklo 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?

The fundemental theorem of calculus is the same for trig functions....

$\int_{a}^{b}f'(x)dx=f(b)-f(a)$

Just evaluate the antiderivative at the end points and subtract.

12. 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.

13. Originally Posted by Meeklo 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.

$\int_{-\pi/3}^{\pi/3}4-\sec^2(x)dx=4x-\tan(x) \bigg|_{-\pi/3}^{\pi/3}=4\left( \frac{\pi}{3}\right) -\tan\left( \frac{\pi}{3}\right)-4\left( \frac{-\pi}{3}\right)+\tan\left( \frac{-\pi}{3}\right)=$
$8\left( \frac{\pi}{3}\right)-2\tan\left( \frac{\pi}{3}\right) \approx 4.91$

14. OK I will go with that.

Thank you very much for your help and insight!