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Math Help - Area Between 2 Graphs (integration)

  1. #1
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    Area Between 2 Graphs (integration)



    The shaded area is the area needed. I can't seem to get the same answer as the book.

    I think the problem is I am not integrating the 'root' 2x correctly.

    Can someone tell how to integrate this?

    Thanks

    (PS. Answer = 2 )
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by r_maths View Post


    The shaded area is the area needed. I can't seem to get the same answer as the book.

    I think the problem is I am not integrating the 'root' 2x correctly.

    Can someone tell how to integrate this?

    Thanks

    (PS. Answer = 2 )
    here
    Attached Thumbnails Attached Thumbnails Area Between 2 Graphs (integration)-area.gif  
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by r_maths View Post

    I think the problem is I am not integrating the 'root' 2x correctly.
    you did in fact integrate this incorrectly. you need to use substitution to do this (that is the standard way--from experience you would know that the integral is half of the answer you got). the way i did it, i didn't have to integrate that. if you're still not sure about how to integrate something like this, say so
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Here's some good stuff to know
    Attached Thumbnails Attached Thumbnails Area Between 2 Graphs (integration)-gened.gif  
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  5. #5
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    Thanks alot!

    ---

    I've another question though.



    I worked out where the graphs intersects.
    0.25^2 - 3x + 8 = 0
    x = 4 or x = 8
    However, how is it possible to know which is the correct one?
    The answer is 26 and 2/3 (26.666...) so x = 4.
    The only way I knew it was 4 is because I looked at the answer, but obviously in the exams
    I will not have the answers to check at. Thanks

    EDIT: Last question, I only worked out half of the graph and times answer by 2, is this ok?
    Last edited by r_maths; April 11th 2007 at 03:34 AM.
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