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Math Help - Midpoint Approx. of Integral

  1. #1
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    Midpoint Approx. of Integral

    I'm not exactly sure how to implement the formula for this.

    \int{x^2 + 1} dx from 0 to 4

    And the question wants MID(2), which I assume means to divide it into two subsections.

    How would I find this?

    So far I have drawn the graph and gotten delta x to be 1. So I have

    f(2 + \frac{2}{2})(1) + f(10 + \frac{2}{2})(1)
     = f(5/2) + f(21/2)
     = (\frac{5}{2})^2 + 1) + ((\frac{21}{2})^2 + 1)

    according to the equation given to me in my notes, which was:

     \int{f(x)} = f(x_1 + \frac{delta x_1} {2}) (delta x_1) + f (x_2 + \frac{delta x_2} {2})(delta x_2)

    So I got 118.5. Is that right?
    Last edited by lysserloo; February 1st 2010 at 07:22 PM.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by lysserloo View Post
    I'm not exactly sure how to implement the formula for this.

    \int{x^2 + 1} dx from 0 to 4

    And the question wants MID(2), which I assume means to divide it into two subsections.

    How would I find this?

    So far I have drawn the graph and gotten delta x to be 1. So I have

    f(2 + \frac{2}{2})(1) + f(10 + \frac{2}{2})(1)
     = f(5/2) + f(21/2)
     = (\frac{5}{2})^2 + 1) + ((\frac{21}{2})^2 + 1)

    according to the equation given to me in my notes, which was:

     \int{f(x)} = f(x_1 + \frac{delta x_1} {2}) (delta x_1) + f (x_2 + \frac{delta x_2} {2})(delta x_2)

    So I got 118.5. Is that right?
    This may sound strange, but could you define MID2? I'm not quite certain what you mean?
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  3. #3
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    MID(2) means n = 2 I believe. Divided into 2 subsections, so like from 0 to 2, then 2 to 4.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by lysserloo View Post
    MID(2) means n = 2 I believe. Divided into 2 subsections, so like from 0 to 2, then 2 to 4.
    So like \frac{1}{b-a}\int_a^b{f(x)}dx?
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  5. #5
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    No it's the equation I gave above, using the midpoint to estimate the integral.

    Well, actually yours may be the same. I'm really not sure.

    Either way, I'm looking for a midpoint approximation, with delta x being
    \frac{b -a}{n} and n = 2 I think.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by lysserloo View Post
    No it's the equation I gave above, using the midpoint to estimate the integral.

    Well, actually yours may be the same. I'm really not sure.

    Either way, I'm looking for a midpoint approximation, with delta x being
    \frac{b -a}{n} and n = 2 I think.
    Oh, I see...So, you want

    \int_0^4f(x)dx=\int_0^4(x^2+1)dx\approx\sum_{i=1}^  2\left[f\left(x_i+\frac{\Delta{x}}{2}\right)\right]{\Delta{x}}

    where x_1=0,x_2=2 and \frac{\Delta{x}}{2}=\frac{4-0}{2}=2?
    Last edited by VonNemo19; February 1st 2010 at 09:17 PM.
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  7. #7
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    Yeah, that looks right.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by lysserloo View Post
    Yeah, that looks right.
    Well then

    \int_0^4f(x)dx\approx{f(2)\Delta{x}+f(4)\Delta{x}}
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