Please help Struggling with finding the area using upper & lower sums

  • Thread starter BuBbLeS01
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  • #1
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Please help!!! Struggling with finding the area using upper & lower sums!!

I can't load the picture on here so I will explain it the best I can...
It is the graph of 1/x and there are 5 subintervals starting at x = 1 and ending at x = 2.
It says to use upper and lower sums to approximate the area of the region. Their widths are equal.

So first I have to find the left and right endpoints I think...but I am still not understanding how to figure them out.
My guess is...
Left: 5(i-1)/n
Right 5i/n
 

Answers and Replies

  • #2
HallsofIvy
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I can't load the picture on here so I will explain it the best I can...
It is the graph of 1/x and there are 5 subintervals starting at x = 1 and ending at x = 2.
It says to use upper and lower sums to approximate the area of the region. Their widths are equal.

So first I have to find the left and right endpoints I think...but I am still not understanding how to figure them out.
My guess is...
Left: 5(i-1)/n
Right 5i/n

Not quite! You have both "n" and "5" in your formula but, if am reading this correctly, n is the number of intervals and i is "stepping" through those intervals: here n= 5. Since y= 1/x is a decreasing function, its highest value on an interval is at the left end of the interval and its lowes value is at the right end. Your region has length 2-1= 1 and dividing it into 5 regions makes each of length 1/5. Because you are given a specific value of n, there should be no "n" in your formula.
 
  • #3
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Oh okay so 1/5 is my width, but then what is my height??
 
  • #4
Gib Z
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Well for the upper sums, the rectangles should go over the graph, so the heights the starting value isn't it? For the lower sums, the rectangles have to be below the graph, so its the smaller value.
 
  • #5
HallsofIvy
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I said before, "Since y= 1/x is a decreasing function, its highest value on an interval is at the left end of the interval and its lowest value is at the right end." Evaluate 1/x at the endpoints of the interval. Of course, the right endpoint of one interval is the left endpoint of the next interval so your sums will differ only at the beginning and end.
 

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