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Thread: Please I need help with this integral calculus...

  1. May 31st 2007, 02:44 AM #1
    White Devil
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    Please I need help with this integral calculus...

    Hello...

    Could you please help me out with this integral calculus... there is a though in this question...
    Attached Thumbnails Attached Thumbnails Please I need help with this integral calculus...-m.jpg  
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  2. May 31st 2007, 02:51 AM #2
    earboth
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    Quote Originally Posted by White Devil View Post
    Hello...

    Could you please help me out with this integral calculus... there is a though in this question...
    Hello,

    I hope I read your problem correctly:

    \int_{-2}^2 x \cdot dx = \left[\frac{1}{2} x^2 \right]_{-2}^2 = \frac{1}{2}(2^2) - \frac{1}{2}((-2)^2) = 0

    But what exactly didn't you understand?
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  3. May 31st 2007, 03:02 AM #3
    White Devil
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    Sorry, I forgot to say the answer is not 0 nor 4....
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  4. May 31st 2007, 03:06 AM #4
    CaptainBlack
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    Quote Originally Posted by White Devil View Post
    Hello...

    Could you please help me out with this integral calculus... there is a though in this question...
    Your question is, evaluate:

    <br /> \int_{-2}^2 x\ dx= \left. \frac{x^2}{2}\right|_{-2}^2 = 2-2 = 0<br />

    Now why are you having problems with this? This question is about as
    simple a definite integral as you are going to see and if you are having
    problems with it you need more help than just being shown the solution.

    RonL
    Last edited by CaptainBlack; May 31st 2007 at 04:32 AM.
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  5. May 31st 2007, 03:10 AM #5
    White Devil
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    The answer is not 0... the teacher said there is a though in it...
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  6. May 31st 2007, 03:22 AM #6
    chogo
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    hey white devil the answer is definitely zero and not anything else
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  7. May 31st 2007, 03:29 AM #7
    White Devil
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    Well the thing is our teacher said he will give 5 marks to who will answer it... and some students did it...
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  8. May 31st 2007, 04:30 AM #8
    Rich B.
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    Hi:

    I am a teacher and such status does not protect me from being wrong. Nor does it do so for your teacher (else, he/she has one terrific contract!).

    Now, picture the graph of y=x on the interval [-4, 4]. As the function is symmetric about the origin, the integral from -4 to 0 is necessarily opposite that from 0 to 4. The sum of two opposites is, of course, zero.

    Regards,

    Rich B.
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  9. May 31st 2007, 04:34 AM #9
    CaptainBlack
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    Quote Originally Posted by White Devil View Post
    The answer is not 0... the teacher said there is a though in it...
    The integral you gave is indeed zero, if your teacher thinks otherwise then
    either they are wrong or you have not asked the question that your teacher
    actually set.

    Can you post what you were asked verbatim?

    RonL
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  10. May 31st 2007, 05:00 AM #10
    ThePerfectHacker
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    Quote Originally Posted by White Devil View Post
    The answer is not 0... the teacher said there is a though in it...
    Tell your teacher to go kill herself.
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  11. May 31st 2007, 05:06 AM #11
    earboth
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    Quote Originally Posted by White Devil View Post
    The answer is not 0... the teacher said there is a though in it...
    Hello,

    it's me again. If the answer isn't zero the only possibility I can think of is that you should calculate the area which is enclosed by y = x and x = -2 and x = 2.

    The straight line with the equation y = x has a zero at x = 0. Thus you have to split the integral into 2 parts:

    \left| \int_{-2}^0 x dx \right| +\left| \int_{0}^2 x dx \right| =\left| \left. \frac{1}{2}x^2 \right|_{-2}^0 \right| +\left| \left. \frac{1}{2}x^2 \right|_{0}^2 \right| = 2 + 2 = 4
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  12. May 31st 2007, 05:46 AM #12
    White Devil
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    earboth you're right I have asked in other forum (Arabic one) and they told me the answer and it's exactly as you did...

    But could you please make it graphical?
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  13. May 31st 2007, 06:21 AM #13
    CaptainBlack
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    Quote Originally Posted by ThePerfectHacker View Post
    Tell your teacher to go kill herself.
    Please consider youself to have been given an infraction for inappropriate
    language

    RonL
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  14. May 31st 2007, 11:36 AM #14
    earboth
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    Quote Originally Posted by White Devil View Post
    earboth you're right I have asked in other forum (Arabic one) and they told me the answer and it's exactly as you did...

    But could you please make it graphical?
    Hello,

    I'm not quite certain what you are asking for. I've attached a diagram showing the area in question:
    Attached Thumbnails Attached Thumbnails Please I need help with this integral calculus...-flaeche_gerade.gif  
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  15. May 31st 2007, 12:09 PM #15
    White Devil
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    Thank you earboth that's what I asked for... Thanks all...
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