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Math Help - Using limits to evaluate integral?

  1. #1
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    Using limits to evaluate integral?

    Use limits to evaluate integral

    y=(x+2) from x=1 to x=4

    I know delta x is 4/n and xi is 4i/n

    my issue is with the substitution

    (4i/n)+2 times (4/n) how would I do this part
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    Re: Using limits to evaluate integral?

    You can't evaluate \displaystyle \int_1^4{x + 2\,dx}?
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    Re: Using limits to evaluate integral?

    I am not sure my textbook just shows in the example using limits. Right now technically I am not learning calculus only an "introduction to calculus" so I have not been taught that yet.

    I have only been taught to integrate using formula such as n(n+1)/2 for 1+2+3+n
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    Re: Using limits to evaluate integral?

    Quote Originally Posted by homeylova223 View Post
    Use limits to evaluate integral

    y=(x+2) from x=1 to x=4

    I know delta x is 4/n and xi is 4i/n

    my issue is with the substitution

    (4i/n)+2 times (4/n) how would I do this part
    If you must use Riemann Sums, first of all, your distance is 3, not 4, so the size of each interval is actually \displaystyle \frac{3}{n}, where n is the number of intervals. This is the length of each of your rectangles.

    What is the width of each of your rectangles?
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    Re: Using limits to evaluate integral?

    Quote Originally Posted by homeylova223 View Post
    Use limits to evaluate integral

    y=(x+2) from x=1 to x=4

    I know delta x is 4/n and xi is 4i/n

    my issue is with the substitution

    (4i/n)+2 times (4/n) how would I do this part
    Learn from every problem you solve...

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    Re: Using limits to evaluate integral?

    if we chop the interval [a,b] into n pieces, each piece has length (b-a)/n. so the k-th out of n subintervals is the interval:

    [a + (b-a)(k-1)/n, a + (b-a)k/n]. in this example a = 1, and b = 4. so we have the intervals:

    [1,1+3/n], [1+3/n, 1+ 6/n],....,[1+(3k-3)/n, 1+3k/n],.....,[1+(3n-3)/n, 4].

    now, to pick a value for xk, we need to decide how we're going to do it. let's go with the right-hand endpoint of each interval

    (since f(x) = x+2 is increasing, this will be a slightly high estimate for the integral).

    this gives us:

    \sum_{k=1}^n f(1+\frac{3k}{n})(\frac{3}{n})

     = \frac{3}{n}\sum_{k=1}^n (\frac{3k}{n}+3) = \frac{3}{n}\sum_{k=1}^n \frac{3k}{n}\ +\ \frac{3}{n}\sum_{k=1}^n 3

     = \frac{9}{n^2}\sum_{k=1}^n k\ +\ 9 = \frac{9(n+1)}{2n} + 9

    now, what we actually want is: \lim_{n \to \infty} (\frac{9(n+1)}{2n} + 9)

    i've done the hard part, now you do the easy part (and convince yourself that this matches your intuitive idea of the area of the region bounded by the x-axis, the lines x = 1 and x = 4, and the graph of f).
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    Re: Using limits to evaluate integral?

    I think I understand know I'll have to try and solve it.

    But what confuses me is when I have (4/n)(4i/n+2) Would I just use distributive property

    Or add it first if I add it first I would need like terms

    And If I did distributive property I dont know if there is a sum for 8/n
    Last edited by homeylova223; July 2nd 2011 at 11:15 AM.
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    Re: Using limits to evaluate integral?

    i'm not sure why you keep fixating on 4. your interval ends at 4, but it starts at 1, so it has length 4-1 = 3. if you chop an interval 3 units long into n pieces, each piece has length 3/n, not 4/n.

    also f(xi) is NOT 4xi/n + 2. the i-th interval (out of our n) goes from 1 + (4-1)(i-1)/n to 1 + (4-1)(i)/n.

    if we pick the right-hand endpoint of this for xi, then xi = 1 + 3i/n (i used k instead of i in my earlier post, its a personal preference).

    f(xi) = (xi) + 2 = (1 + 3i/n) + 2 = 3i/n + 3. look, no 4's. now if we sum:

    \sum_{i=1}^n (3i/n + 3)(3/n), we can take the factor 3/n out of the sum (by the distributive law).

    that leaves us with:

    (3/n)[\sum_{i=1}^n\ i/n + \sum_{i=1}^n 3]

    the second summation is just n 3's, so we get:

    (3/n)[\sum_{i=1}^n\ i/n +\ 3] = (3/n)\sum_{i=1}^n\ i/n + (3/n)(3n)

    in the second sum, the n's cancel, leaving (3)(3) = 9. so our sum is 9 plus something, the something being:

    (3/n)\sum_{i=1}^n\ i/n. again we can take the factor of 1/n out in front, so we have:

    [(3/n^2)\sum_{i=1}^n\ i\] +\ 9

    now surely you know that \sum_{i=1}^n\ i = n(n+1)/2,

    so our formula becomes (3/n^2)(n(n+1))/2 + 9 = (3/n)((n+1)/2) + 9,

    which is the same as (3/2)((n+1)/n) + 9, as in my previous post.

    i understand that calculating riemann sums is a tedious process, which requires careful attention to the algebra, but there is no easy way around it.
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    Re: Using limits to evaluate integral?

    Hmm your explanation is crisp and clear like 7 up. But I wonder how did you get +3 into (3/n)(3n).
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    Re: Using limits to evaluate integral?

    Quote Originally Posted by homeylova223 View Post
    Hmm your explanation is crisp and clear like 7 up. But I wonder how did you get +3 into (3/n)(3n).
    Your question makes no sense. (3/n)(3n) is equal to 9, which is what Deveno said!

    If you are wondering where the (3n) came from in (3/n)(3n), you're expected to know the basic property that \sum_{i=1}^n 3 = 3n.
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