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Math Help - Integration With Simpson's Rule

  1. #1
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    Integration With Simpson's Rule

    Hi Everone,

    I have a problem that I am struggling to solve, I am very new to intergration and until now I have been coping OK. I am strugling to identify the significance of the 2 before the intergral sign, Please can someone advise how I tackle this problem.. Thanks,

    2 \int 470 di

    I am sorry but I dont know how to right the limits on here but the upper limit is 1.0 and the lower is 1.00
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    Quote Originally Posted by duckegg View Post
    Hi Everone,

    I have a problem that I am struggling to solve, I am very new to intergration and until now I have been coping OK. I am strugling to identify the significance of the 2 before the intergral sign, Please can someone advise how I tackle this problem.. Thanks,

    2 \int 470 di

    I am sorry but I dont know how to right the limits on here but the upper limit is 1.0 and the lower is 1.00
    I think the 2 is just a multiplicative factor out in the front but I get the impression that the integral as written is not what you intended. What should that 470 be? And are you integrating over "i"?

    -Dan

    Edit: Unless you are integrating the constant 470 over i?
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    Quote Originally Posted by topsquark View Post
    I think the 2 is just a multiplicative factor out in the front but I get the impression that the integral as written is not what you intended. What should that 470 be? And are you integrating over "i"?

    -Dan

    Edit: Unless you are integrating the constant 470 over i?
    Dan frist of all thank you for the very quick reply... Here is the full problem to help explain my difficulty..

    In elctrical theory we may plot voltage, V vertically against current, I(A) horizontally. The area under such curve is referred to as power, P(W) ehich can be determined by the formula...

    2 \int Vdi With lower limit (a) and upper limit (b)

    For a resistor of 470 \Omega where V = 470I determine the power associated with an increase in current from 1mA to 1.08mA.
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    Quote Originally Posted by duckegg View Post
    Dan frist of all thank you for the very quick reply... Here is the full problem to help explain my difficulty..

    In elctrical theory we may plot voltage, V vertically against current, I(A) horizontally. The area under such curve is referred to as power, P(W) ehich can be determined by the formula...

    2 \int Vdi With lower limit (a) and upper limit (b)

    For a resistor of 470 \Omega where V = 470I determine the power associated with an increase in current from 1mA to 1.08mA.
    Okay, so
    2 \int 140I~dI

    So yes, the 2 is merely a multiplicative factor. ie. multiply your answer to the integration by 2.

    Next question: Your proposed integration limits are 1.0 = 1.00? That would result in
    2 \int_{1.00}^{1.0} 140I~dI = 0

    since integration over equal limits is 0. I don't think this is what you meant either?

    -Dan
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    Hi Sorry,

    The upper limit should be 1.08 and the lower 1.00
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    Also please could you explain where you got 140 from? Thanks, Chris
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by duckegg View Post
    Also please could you explain where you got 140 from? Thanks, Chris
    You yourself said that P = 2 \int V~dI and that V = 140 I. That's where the 140 comes from. So you need to calculate 2 \int 140I~dI.

    -Dan
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    Sorry the actual value I put was 470. I now get the answer to be 75.2 is this correct?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by duckegg View Post
    Sorry the actual value I put was 470. I now get the answer to be 75.2 is this correct?
    Sorry about the 470 thing.

    2 \int_{1.00}^{1.08}470I~dI = 2 \cdot 470 \cdot \frac{I^2}{2} \bigg |_{1.00}^{1.08} = 470*(1.08^2 - 1.00^2) = 78.208

    -Dan
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    Excellent thats great Dan, Thank You. Now the only thing I have to do is proove it using Simpson's Rule! Breaking it down into 6 Equal Spacings. How are you with simpsons rule. I've had a go but cant get the same answer as above. Is this formula correct...

    P=[1.08-1.00] Divided By 6 My answer to this comes out at 0.013 which cause me a problem when entering the values into a table. I know that values for x must be even values??
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    Quote Originally Posted by duckegg View Post
    Excellent thats great Dan, Thank You. Now the only thing I have to do is proove it using Simpson's Rule! Breaking it down into 6 Equal Spacings. How are you with simpsons rule. I've had a go but cant get the same answer as above. Is this formula correct...

    P=[1.08-1.00] Divided By 6 My answer to this comes out at 0.013 which cause me a problem when entering the values into a table. I know that values for x must be even values??
    The integral is approximated by
    \int_a^b f(x)~dx \approx \frac{\Delta x}{3}(y_0 + 4y_1 + 2y_2 + 4y_3 + \text{...}~4y_{2n - 1} + y_{2n} )

    The spacings are given by
    \Delta x = \frac{b - a}{2n}

    and
    y_k = f(a + k \Delta x)

    So if you are doing 6 intervals (n = 3), then
    \Delta I = \frac{1.08 - 1.00)}{2 \cdot 3} = 0.013333

    (I switched the independent variable to I as it is in your problem.)

    So your table looks like:
    \begin{array}{|c|c|} \hline I & 470I \\ \hline 1.000000 & 470.000000 \\ 1.013333 & 476.26651 \\ 1.026666 & 482.65502 \\ \hline \end{array}

    etc.

    The sum you need to do then, is
    \int_{1.00}^{1.08} 470I~dI \approx \frac{0.013333}{3}(470.000000 + 4(476.26651) + 2(482.65502) + ~\text{...} )

    And don't forget to multiply this by 2 as stated in the original problem. I get 78.205970
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    Hi Thanks for the help Dan,

    However correct me if I'm wrong but the table doesn't seem to work.

    I.e. 470 x 1.013 = 476.11 not 476.26 like in your table,

    Am I missing something? Thanks,

    Chris
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