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Math Help - Reimann's sums

  1. #1
    Member akhayoon's Avatar
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    Reimann's sums

    \int_{1}^{3}x^2-x+1~dx

    I have to find this integral using reimann's sums

    usually what I do si subtract 3-1/n to get 2/n

    and that gives me 2i/n which I then sub into the polynomial and multiply by 2/n to give me the answer(using limits and sums ofcourse)...but for this one it's not working so I'm guessing I shouldn't be using 2i/n???
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
    \int_{1}^{3}x^2-x+1~dx

    I have to find this integral using reimann's sums

    usually what I do si subtract 3-1/n to get 2/n

    and that gives me 2i/n which I then sub into the polynomial and multiply by 2/n to give me the answer(using limits and sums ofcourse)...but for this one it's not working so I'm guessing I shouldn't be using 2i/n???
    no, it is not 2i/n

    you have: x_0 = 1, x_1 = 1 + \frac 2n, x_2 = 1 + \frac 4n \cdots x_i = 1 + \frac {2i}n

    and of course, \int_a^b f(x)~dx = \lim_{n \to \infty} \sum_{i = 0}^{n} f(x_i) \Delta x, where \Delta x = \frac {b - a}n
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  3. #3
    Member akhayoon's Avatar
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    thanks but I don't understand how you got

    x_1=1+\frac{2}{n} or

    x_2=1+\frac{4}{n}

    could you please show me how you got these?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
    thanks but I don't understand how you got

    x_1=1+\frac{2}{n} or

    x_2=1+\frac{4}{n}

    could you please show me how you got these?
    remember, the x_i's are the points on the x-axis that are \Delta x apart.

    so if x_0 = a

    then x_1 = a + \Delta x .....since we move a distance of \Delta x to the right to go from x_0 to x_1

    and x_2 = x_1 + \Delta x

    and x_3 = x_2 + \Delta x

    and so on.

    here, \Delta x = \frac 2n

    so, x_0 = 1

    and x_1 = 1 + \frac 2n

    and x_2 = x_1 + \Delta x = 1 + \frac 2n + \frac 2n = 1 + \frac 4n

    and x_3 = x_2 + \Delta x = 1 + \frac 4n + \frac 2n = 1 + \frac 6n

    .
    .
    .

    x_i = 1 + \frac {2i}n
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