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I need serious help with this problem, it's about area and distance

  1. Apr 26, 2007 #1
    1. The problem statement, all variables and given/known data

    by reading values from the given graph of f, use five rectangles to find a lower estimate and an upper estimate for the area under the given graph of from x=0 to x=10. in each case sketch the rectangles that you use.



    3. The attempt at a solution
    since their are there are 5 rectangles, the width was 2 for each of them

    and when i tried to figure out the lower and upper estimate i did this

    L5= 2*O^2+ 2*2^2 ..... 2 * 8^2 =128

    and the answer in the back of the book is 40

    and for Right end= 2*2^2 ..... 2*10^2= 440

    and the answer is 52
     
  2. jcsd
  3. Apr 26, 2007 #2
    i already did sketched the rectangle thing, and it doesn't tell me what type of graph it is, it just tells me y= f(x)
     
  4. Apr 26, 2007 #3

    Dick

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    The area of a rectangle is length*width. There is no 'square' in the formula.
     
  5. Apr 26, 2007 #4
    ok, so then i dont have to square them,
     
  6. Apr 26, 2007 #5
    i was looking at an example that was squaring them, no wonder, thanks man, i didn't think about the y= f(X)
     
  7. Apr 26, 2007 #6
    ok so what do i do then
     
  8. Apr 26, 2007 #7
    i didnt square the left end points and i got 40
    but i did the same for the right end and i got 60
     
  9. Apr 26, 2007 #8

    Dick

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    You must be able to determine the values of f(0), f(2) etc from your graph. Those are the heights. The widths are all 2.
     
  10. Apr 26, 2007 #9
    how do i do that, what formula do i use
     
  11. Apr 26, 2007 #10

    Dick

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    If you are going to get a numerical answer like 52 then you must be able to find f(0) etc. It's the y value of the graph when x=0 etc.
     
  12. Apr 26, 2007 #11
    i did it, i multiplied 2 * the hieght, for every one
     
  13. Apr 26, 2007 #12

    Dick

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    And got 52 for one of them?
     
  14. Apr 26, 2007 #13
    yup, i got 53, but then i took a closer look at the heights and got 52
     
  15. Apr 26, 2007 #14
    thanks for the help
     
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