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Approximating Areas Under Curves

  1. Apr 22, 2012 #1
    Good morning!

    I'm having a bit of a hard time wrapping my mind around this concept. The part that seems to be troubling me the most is where Riemann Sums and sigma notation come in. I can't seem to find explanations online that help me better understand, so hopefully someone here could help?

    The textbook that I am using is Briggs and Cochran's Calculus, section 5.1 specifically. For anybody that has this book, I have tried to follow example #5 as it seemed like it would be the most helpful. I have no clue what exactly they've done. The problem asks you to evaluate the left, right, and midpoint Riemann sums of f(x)= x^3 + 1 between a=0 and b=2 using n=50 subintervals. I can't figure out what to do once I have my Δx. The book just leaves me lost.
     
  2. jcsd
  3. Apr 22, 2012 #2
    First, divide the interval on the x-axis between a = 0 and b = 2 into 50 equally large subintervals. The size of each subinterval is delta-x. You are then to erect a rectangle over each interval up to the height of the curve above each interval, then add up the areas of the 50 rectangles to approximate the area under the curve.
    The problem is that the curve has many different values above each subinterval. One way to pick a value is to pick the leftmost value in each interval for the height of each rectangle. Another is to pick the rightmost value, and a third is to pick the value above the midpoint of each subinterval. It is a good idea to draw a picture of the curve and the rectangles to see what to do.
    To get the leftmost value as the height of each rectangle, first find the leftmost x-value for each interval. Since we divided the interval between 0 and 2 into 50 equally large intervals, each interval has a length of delta-x = 2/50 = 1/25. For the first interval, the leftmost value is a, which is 0. The second interval starts at 0 + 1/25. The third interval starts at 0 + 2/25. And so forth up to 0 + 49/25 for the last interval. So the leftmost value in the nth interval is (n-1)/25, where n goes from 1 to 50.
    Now we need to get the height of each rectangle using this leftmost x-value. The height of the curve above the leftmost point of each interval is given by the function. For each value of n, we let x = (n - 1)/25 to get [(n-1)/25]^3 + 1 as the height of the curve above that particular x-value.
    To get the area of the nth rectangle, we multiply its height by its width. Its width is constant, the width of each subinterval: delta-x. The height of the nth rectangle using the leftmost point is [(n-1)/25]^3 + 1. Put together, the area of the nth rectangle is then (1/25)([(n-1)/25]^3 + 1). We then add up all these areas from n = 1 to n = 50 to get the approximation of the area under the curve using 50 rectangles with the leftmost value of each subinterval as the rectangle height values.
    You can do a similar analysis to figure out how to get the midpoint and right sums.
     
  4. Apr 22, 2012 #3
    Thanks for the quick reply! :)

    I really think the main problem I'm having is with using Riemann Sums (using sigma notation).

    When n is a large number, I'm lost. I've tried to follow this text and watch videos, but nobody seems to explain how to find these sums in sigma notation.
     
    Last edited: Apr 22, 2012
  5. Apr 22, 2012 #4
    The notation is not a method of finding sums. It is just a short way of writing a long sum. For example, the last thing I mentioned above is to add up the 50 numbers of the form (1/25)([(n-1)/25]^3 + 1). To do that, evaluate that number for each value of n, from n = 0 to n = 50, and add those numbers up. That is the sum {(1/25)([(1-1)/25]^3 + 1} + {(1/25)([(2-1)/25]^3 + 1} + {(1/25)([(3-1)/25]^3 + 1} + {(1/25)([(4-1)/25]^3 + 1} + ... + {(1/25)([(49-1)/25]^3 + 1} + {(1/25)([(50-1)/25]^3 + 1}.
    In sigma notation, the same thing is written as
    [tex]\sum_{n=1}^{50} \left(\left(\frac{n-1}{25}\right)^3 + 1\right)\left(\frac{1}{25}\right)[/tex]
    It does not help you evaluate it in any way; it just saves you from having to write out all 50 terms when algebraically manipulating the sum. For example, you can factor out the delta-x, which is 1/25, since it is a factor of every term.
    [tex]\sum_{n=1}^{50} \left(\left(\frac{n-1}{25}\right)^3 + 1\right)\left(\frac{1}{25}\right) = \frac{1}{25}\sum_{n=1}^{50} \left(\left(\frac{n-1}{25}\right)^3 + 1\right)[/tex]
    If you write the sum out term by term, you can see that we will be adding up 50 1's, so we may as well do that now:
    [tex]\frac{1}{25}\left(\sum_{n=1}^{50} \left(\left(\frac{n-1}{25}\right)^3 + 1\right)\right) = \frac{1}{25}\left(\left(\sum_{n=1}^{50} \left(\frac{n-1}{25}\right)^3\right) + 50\right) = \frac{1}{25}\left(\sum_{n=1}^{50}\left(\frac{n-1}{25}\right)^3\right) + 2[/tex]
    We can now factor out that (1/25)^3:
    [tex]\frac{1}{25}\left(\sum_{n=1}^{50}\left(\frac{n-1}{25}\right)^3\right) + 2 = \frac{1}{25^4}\left(\sum_{n=1}^{50}(n-1)^3\right) + 2[/tex]
    We can get a more familiar sum by letting u = n - 1:
    [tex]\frac{1}{25^4}\left(\sum_{n=1}^{50}(n-1)^3\right) + 2 = \frac{1}{25^4}\left(\sum_{u=0}^{49}u^3\right) + 2[/tex]
    The sum of the first n cubes is (n(n+1)/2)^2, so we get:
    [tex]\frac{1}{25^4}\left(\sum_{u=0}^{49}u^3\right) + 2 = \frac{1}{25^4}\left(\frac{49(49+1)}{2}\right)^2 + 2[/tex]
    It is just factoring and breaking a sum up into more manageable pieces. The only thing the notation did was ensure we did not have to write 50 or more terms on each line. You do have to know a few elementary sums if you want to avoid using a calculator for summing. Graphing calculators will usually let you enter a sum directly and they will evaluate it for you. Without a calculator, if you run across a sum you do not recognize, you might only be able to evaluate it the old fashioned way, by evaluating each term and adding them up.
     
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