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Math Help - Definition of the integral question (Riemann sums?)

  1. #1
    s3a
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    Definition of the integral question (Riemann sums?)

    The following sum

    is a right Riemann sum for the definite integral
    where = ? = 3
    and = ? = sqrt(9-x^2)

    The part of the question I'm stuck on is:
    The limit of these Riemann sums as is = ?

    I know that I need to evaluate:
    lim(n->inf) sum from 0 to n of sqrt(9-(3i/n)^2) but after scaring the 3i/n part, I'm stuck.

    Any help would be GREATLY appreciated!
    Thanks in advance!

    P.S.
    Does this whole question involve a topic called "Riemann sums?"
    Last edited by s3a; May 20th 2010 at 12:56 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Are you supposed to obtain the limit of the Riemann sum OR integrate?
    And n goes to infinity not x.
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  3. #3
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    Quote Originally Posted by s3a View Post
    The following sum

    is a right Riemann sum for the definite integral
    where = ? = 3
    and = ? = sqrt(9-x^2)

    The part of the question I'm stuck on is:
    The limit of these Riemann sums as is = ?

    I know that I need to evaluate:
    lim(x->inf) sum from 0 to n of sqrt(9-(3i/n)^2) but after scaring the 3i/n part, I'm stuck.

    Any help would be GREATLY appreciated!
    Thanks in advance!

    P.S.
    Does this whole question involve a topic called "Riemann sums?"
    Well, the whole point is that since that sum is a Riemann sum for the integral \int_0^3 \sqrt{9- x^2} dx its limit as n goes to infinity must be the integral!

    And you don't even have to do an integral. If y= \sqrt{9- x^2}, then x^2+ y^2= 9 except that y\ge 0- the upper half of a circle of radius 3 with center at the origin. And since the integral is from x= 0 to x= 3 rather than -3 to 3, that integral is the area of 1/4 of a circle of radius 3.
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  4. #4
    s3a
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    Quote Originally Posted by matheagle View Post
    Are you supposed to obtain the limit of the Riemann sum OR integrate?
    And n goes to infinity not x.
    Ya n->inf not x (sorry force of habit). And I'm supposed to evaluate the limit of the Riemann sum for this. How do I do this algebraically?
    Last edited by s3a; May 20th 2010 at 02:17 PM.
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    Quote Originally Posted by s3a View Post
    Ya n->inf not x (sorry force of habit). And I'm supposed to evaluate the limit of the Riemann sum for this. How do I do this algebraically?
    You have been given that answer! Read below.
    Quote Originally Posted by HallsofIvy View Post
    Well, the whole point is that since that sum is a Riemann sum for the integral \int_0^3 \sqrt{9- x^2} dx its limit as n goes to infinity must be the integral!
    And you don't even have to do an integral. If y= \sqrt{9- x^2}, then x^2+ y^2= 9 except that y\ge 0- the upper half of a circle of radius 3 with center at the origin. And since the integral is from x= 0 to x= 3 rather than -3 to 3, that integral is the area of 1/4 of a circle of radius 3.
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  6. #6
    s3a
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    Oh I misread that thinking he wanted me to do the integral. But still, is there not an algebraic way? (I did the graphing way)
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  7. #7
    MHF Contributor matheagle's Avatar
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    let x=3\sin u
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  8. #8
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    Quote Originally Posted by s3a View Post
    Oh I misread that thinking he wanted me to do the integral. But still, is there not an algebraic way? (I did the graphing way)
    What do you mean by an "algebraic way" (and, for that matter, what is the "graphing way")?

    As I told you, that is 1/4 of a circle of radius 3. Its area is (1/4)\pi (3^2).
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