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Math Help - integral

  1. #1
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    integral

    can someone help me do this
    the intergral, square root of 1 + 9x^4 dx
    i can't seem to figure it out
    thanks
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  2. #2
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    It might be an elliptic integral, were you working with ellipse when you got it?
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  3. #3
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    Quote Originally Posted by nertil1
    can someone help me do this
    the intergral, square root of 1 + 9x^4 dx
    i can't seem to figure it out
    thanks
    Not an elementary integral. The attachment shows what QuickMath
    has to say about it, together with some explanatory text from
    functions.wolfram.com - though how the -ve modulus is supposed to
    be interpreted I don't know

    RonL
    Attached Thumbnails Attached Thumbnails integral-gash.jpg  
    Last edited by CaptainBlack; July 29th 2006 at 10:40 PM.
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  4. #4
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    No, I was actually trying to find the length of the function f(X) = x^3 from 1 to 4. The formula that was in my calc book said to do this I have to take the derivative of the function first then add 1 and then take the square root of the whole thing. then integrate it
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  5. #5
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    Quote Originally Posted by nertil1
    No, I was actually trying to find the length of the function f(X) = x^3 from 1 to 4. The formula that was in my calc book said to do this I have to take the derivative of the function first then add 1 and then take the square root of the whole thing. then integrate it
    Actually s = \int_a^b \, dx \sqrt{1 + \left ( f'(x) \right ) ^2}. You applied it right, but explained it wrong.

    -Dan
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  6. #6
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    Quote Originally Posted by nertil1
    No, I was actually trying to find the length of the function f(X) = x^3 from 1 to 4. The formula that was in my calc book said to do this I have to take the derivative of the function first then add 1 and then take the square root of the whole thing. then integrate it
    Ah! Then the book wanted you to approximate the integration. That can be done easily, if that is what you want tell us and we solve it for you.
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  7. #7
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    I used the midpoint approximation with 10 subintervals. You could try trapezoid, right, left, Simpson,etc.


    \frac{4-1}{10}=\frac{3}{10}

    Code:
        x                   
    
    
    
    
    \sqrt{1+9x^{4}}
    ------             ----------------
    
    1.15                            4.091584
    1.45                            6.38629
    1.75                            9.241762
    2.05                            12.647097
    2.35                            16.597652
    2.65                            21.091220
    2.95                            26.126645
    3.25                            31.703275
    3.55                            37.820723
    3.85                            44.478743
                                     --------------
                                  (.3)(210.184977)=63.055493
    This is very close to the actual of 63.12410226.

    Of course, the more intervals the greater the accuracy.
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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  8. #8
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    i think there's supposed to be a x^3 somewhere so you can use u=1+9x^4 or maybe not
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  9. #9
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    is there a way to evaluate the integral to get the precise answer or can you only estimate it?
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  10. #10
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    Quote Originally Posted by nertil1
    is there a way to evaluate the integral to get the precise answer or can you only estimate it?
    Let me try to explain you.

    The [I]natural logarithm function[I] is definied as,
    \ln x=\int_1^x \frac{dx}{x} ,x>0.

    Note, if you never made up such a function there is was no way to evaluated for example,
    \int_0^1 \frac{1}{x+1} dx cuz it gives,
    \ln 2. So by introducting the natural logaithm function we can give precise values of certain integral.

    However, this antiderivative cannot be expressed in regular function (like the natural logarithm so we introduce it) unless we introduce another class of function called "elliptic integral" (not to be confused with elliptic curves and elliptic functions). They enable us to express this value.

    In reality, all you use anyways is an approximation so there is no harm in using the trapezodial rule.
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  11. #11
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    Quote Originally Posted by c_323_h
    i think there's supposed to be a x^3 somewhere so you can use u=1+9x^4 or maybe not
    If there were then substitute would be the trick.

    The problem in most arc length problems is that the functions produce integral of non-elementray functions. When the most basic ones like the one here.
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