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Math Help - Real Analysis - Riemann Integral

  1. #1
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    Real Analysis - Riemann Integral

    Hey, I've got a set of questions concerning the Riemann Integral. It's especially difficult since I was absent for all of the classes concerning the topic. Any help would be greatly appreciated:

    Consider f(x) = 2x + 1 over the interval [1,3]. Let P be the partition consisting of the points {1, 3/2, 2, 3}.

    a. Compute L(f,P); U(f,P); and U(f,P) - L(f,P)

    b. What happens to the value of U(f,P) - L(f,P) when we add the point 5/2 to the partition?

    c. Find a partition P' of [1,3] for which U(f,P') - L(f,P') < 2.
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  2. #2
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    ajj86 - I've closed all your similar threads after this one. It isn't polite to list out sequential problems concerning the same topic one right after the other. You need to show some effort on this problem and perhaps it will shed light on the others. I'll unlock the other threads after this one gets attention.
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  3. #3
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    Quote Originally Posted by ajj86 View Post
    Hey, I've got a set of questions concerning the Riemann Integral. It's especially difficult since I was absent for all of the classes concerning the topic. Any help would be greatly appreciated:

    Consider f(x) = 2x + 1 over the interval [1,3]. Let P be the partition consisting of the points {1, 3/2, 2, 3}.
    So you are dividing the interval on the x-axis from 1 to 3 into 3 subintervals, 1 to 3/2, 3/2 to 2, and 2 to 3. The first interval has length 3/2- 1= 1/2, the second interval has length 2- 3/2= 1/2 and the third interval has length 3- 2= 1.

    a. Compute L(f,P);
    "L" for "Lower". Always use the lowest value of f on each interval. Since f(x)= 2x+ 1 is an increasing function, the lowest value of f on any interval is on the left. The lowest value on the the interval from 1 to 3/2 is at x= 1 and is f(1)= 2(1)+1= 3. The area of the rectangle is base*height= (1/2)(3).
    The lowest value on the interval from 3/2 to 2 is at x= 3/2 and is 2(3/2)+ 1= 4. The area of the rectangle is base*height= (1/2)(4).
    The lowest value on the interval from 2 to 3 is at x= 2 and is 2(2)+ 1= 5. The area of the rectangle is (1)(5).

    The area of all rectangles, and L(f,P), is their sum: (1/2)(3)+ (1/2)(4)+ (1)(5).

    U(f,P);
    "U" for "upper". Always use the highest value of f on each interval. Since f is an increasing interval the highest value on any interval is on the right. The highest value on the interval from 1 to 3/2 is at x= 3/2 and is 2(3/2)+ 1= 4. The area of the rectangle is base*height= (1/2)(4).
    The highest value on the interval from 3/2 to 2 is at x= 2 and is 2(2)+ 1= 5. The area of the rectangle is (1/2)(5).
    The highest value on the interval from 2 to 3 is at x= 3 and is 2(3)+ 1= 7. The area of the rectangle is (1)(7).

    The area of all the rectangles, and U(f,P), is their sum: (1/2)(4)+ (1/2)(5)+ (1)(7).

    and U(f,P) - L(f,P)
    Well, now, that's just arithmetic, isn't it?

    b. What happens to the value of U(f,P) - L(f,P) when we add the point 5/2 to the partition?
    The best way to answer this is to repeat part a, actually adding 5/2 to the partition so you have 4 intervals, from 1 to 3/2, 3/2 to 2, 2 to 5/2, and 5/2 to 3. Notice that you can use what you did before from 1 to 3/2 and from 3/2 to 2. The interval from 2 to 3 is divided into two intervals: from 2 to 5/2, the Lower value is at 2 as before but now the base has length 1/2: (1/2)(5). The Lower value on the interval from 5/2 to 3 is at 5/2 and has value 2(5/2)+ 1= 6. Instead of having (1)(5) we have (1/2)(5)+ (1/2)(6) which is slightly larger.
    L(P) now is (1/2)(3)+ (1/2)(4)+ (1/2)(5)+ (1/2)(6), again slightly larger than L(P) before.

    For U(p) we look at the right sides and have f(5/2)= 2(5/2)+ 1= 6 and f(3)= 2(3)+ 1= 7 so U(P) is now (1/2)(4)+ (1/2)(5)+ (1/2)(6)+ (1/2)(7). Of course (1/2)(6)+ (1/2)(7) is smaller than (1/2)(7)+ (1/2)(7)= 1(7) that we had before so this is slightly smaller.

    The result is that U(p) is smaller, L(P) is higher so U(p)- L(P) is smaller. IF, taking more and more intervals, U(P)- L(P) goes to 0,that is that U(P) and L(P) become the same, THEN the function is "integrable" and that common value is the integral.

    The point of taking the "lowest" and "highest" values is to trap the graph itself between them: for every P, U(P) is larger than the actual "area under the graph" and L(P) is less than the actual "area under the graph". If those two go to the same thing, then their common value must be the area under the curve!

    c. Find a partition P' of [1,3] for which U(f,P') - L(f,P') < 2.
    tedious but not difficult. try more and more intervals.
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  4. #4
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    Thank You

    Wow, HallsofIvy, this response is extremely detailed. Many thanks!
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