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Unsure about these two

  1. Aug 26, 2004 #1

    Having trouble with these two problems:

    1. If f(x) = ln(x)/x (3 <= x <= 10), find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

    My answer:
    Change in x = (10 - 3)/n = 7/n, so:

    (7/n)[Sigma i = 0 to n](ln(7/n)/(7/n))

    I feel like this isn't right for some reason.

    2. Determine a region whose area is equal to:

    lim as n->infinity of (Sigma i = 0 to n)(4pi/4n)(tan (i*pi/4n))

    pi/4n is constant so that is the change in x.

    Don't know how to go any further here...

    And help appreciated.
  2. jcsd
  3. Aug 27, 2004 #2


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    Science Advisor

    1. One obvious point- the quantity inside the sum, ln(7/n)/(7/n) is a constant, it doesn't depend on i! Think carefully about what you are doing: You have divided the interval from 3 to 10 into n intervals, each of length 7/n. If x is the left hand endpoint, then the function value there is ln(x)/x so that is what you are using as the height of the rectangle. The area of each small rectangle is (7/n)(ln(x)/x) and x is NOT 7/n: that's the change from one x to another. In fact, x= 3+ (7/n)i ("3+ " because you are starting at 3, "(7/n)i" because you are move 7/n for each increase in i).

    2. Take x=0 as the left endpoint to simplify the calculations. Since the only dependence on i is "i*pi/4n", it appears that you are using a step of pi/4n. That tells us two things:(a) the total change is from 0 to n(pi/4n)= pi/4 so the region must run from 0 to pi/4, (b) the base of each rectangle is pi/4n so the height mujst be the
    4 tan(i*pi/4n) part and the function itself must be 4 tan(x).
    The region you want is the region bounded on the top by y= 4 tan(x), on the bottom by y=0, on the left by x=0, and on the right by x= pi/4.
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