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

  1. #1
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    Could 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!
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    Quote Originally Posted by Meeklo Braca View Post
    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!
    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
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  3. #3
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    Quote Originally Posted by Meeklo Braca View Post
    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!
    A = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 4 - \sec^2{x} \, dx<br />

    integrate and evaluate.
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  4. #4
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    I dont know how to integrate that
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    Quote Originally Posted by Meeklo Braca View Post
    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?
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    i dont know the anti derivative of sec^2x
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    Quote Originally Posted by Meeklo Braca View Post
    i dont know the anti derivative of sec^2x
    what is the derivative of \tan{x} ?
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    I gotcha, So I have A = x^2 -tan x, Where do I go from there?
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    Quote Originally Posted by Meeklo Braca View Post
    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.
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  10. #10
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    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?
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    Quote Originally Posted by Meeklo Braca View Post
    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.
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    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.
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    Quote Originally Posted by Meeklo Braca View Post
    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
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  14. #14
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    OK I will go with that.

    Thank you very much for your help and insight!
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