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Math Help - How do you calculate a Riemann sum involving square roots?

  1. #1
    s3a
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    How do you calculate a Riemann sum involving square roots?

    I know how to calculate Riemann sums however whenever there is a square root involved with the sigma notation, I am completely stuck (with any problem that has a square root).

    What I did get, however, is that b = 5 and that the function f(x) is f(x) = sqrt(25 - x^2). The question is #2 on the "temp.pdf" file. My (limited) work is attached on the "mywork.pdf" file.

    Any help would be GREATLY appreciated!
    Thanks in advance!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by s3a View Post
    I know how to calculate Riemann sums however whenever there is a square root involved with the sigma notation, I am completely stuck (with any problem that has a square root).

    What I did get, however, is that b = 5 and that the function f(x) is f(x) = sqrt(25 - x^2). The question is #2 on the "temp.pdf" file. My (limited) work is attached on the "mywork.pdf" file.

    Any help would be GREATLY appreciated!
    Thanks in advance!
    From context we see that b=5, \Delta{x}=\frac{b-a}{n}=\frac{5}{n}, and x_i=0+i\Delta{x}=\frac{5i}{n}.

    So, we know that the limit of the sum is equal to by definition the definite integral from a to b, or

    \lim_{n\to\infty}\sum_{i=1}^n\sqrt{25-x_i^2}\Delta{x}=:\int_0^5\sqrt{25-x^2}dx.

    Can you integrate?

    EDIT: By the way, y=f(x)=\sqrt{25-x^2}\Longrightarrow{x^2+y^2=(5)^2} which is obviously a circle with radius r=5. So, the easiest way to do this is to compute A=\frac{1}{4}\pi{r}^2
    Last edited by VonNemo19; January 30th 2010 at 02:26 PM. Reason: forgot a square
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  3. #3
    s3a
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    I understood the area way and got the right answer but I'd also appreciate if you could tell me how to integrate the square root. Yes, I do know how to integrate (apart from square roots which I am having trouble with).
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  4. #4
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    Quote Originally Posted by s3a View Post
    I understood the area way and got the right answer but I'd also appreciate if you could tell me how to integrate the square root. Yes, I do know how to integrate (apart from square roots which I am having trouble with).
    trig substitution ... x = 5\sin{\theta}<br />
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by s3a View Post
    I understood the area way and got the right answer but I'd also appreciate if you could tell me how to integrate the square root. Yes, I do know how to integrate (apart from square roots which I am having trouble with).
    Try letting x=5\sin\theta. This implies that dx=5\cos\theta{d}\theta. Then our integral becomes

    \int\sqrt{25-(5\sin\theta)^2}5\cos\theta{d}\theta

    =5\int\sqrt{25(1-\sin^2\theta)}\cos\theta{d}\theta

    =25\int\sqrt{1-\sin^2\theta}\cos\theta{d}\theta

    =25\int\sqrt{\cos^2\theta}\cos\theta{d}\theta

    =25\int{\cos\theta}\cos\theta{d}\theta

    =25\int\cos^2\theta{d}\theta

    Can you finish?
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  6. #6
    s3a
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    I don't know how to replace dx. My teacher didn't teach that yet. In fact, he said not to worry about it for now. Also if this is a "substitution rule," I noticed in my book that that is the next topic therefore I did not cover it so I am not understanding you guys. But, if I have to know something I didn't learn yet to do this properly then I was probably expected to use the "half circle's area" method.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by s3a View Post
    I don't know how to replace dx. My teacher didn't teach that yet. In fact, he said not to worry about it for now.
    Then tell him not to assign freakin problems that require it. Where do you go to school?
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