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

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In summary, the conversation is about using upper and lower sums to approximate the area of a region on a graph of 1/x, with 5 subintervals of equal width. The height of the rectangles for upper and lower sums is determined by evaluating 1/x at the left and right endpoints of each interval.
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BuBbLeS01
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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
 
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  • #2
BuBbLeS01 said:
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
Oh okay so 1/5 is my width, but then what is my height??
 
  • #4
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
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.
 

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

What is the difference between upper and lower sums when finding the area?

The upper sum is an approximation of the area under a curve or within a region that is calculated by using the maximum value of the function over each subinterval. The lower sum is also an approximation, but it uses the minimum value of the function over each subinterval. The difference between the two sums can help determine the accuracy of the approximation.

How do you determine the number of subintervals to use when finding the area?

The number of subintervals is typically determined by the desired level of accuracy. A larger number of subintervals will result in a more accurate approximation of the area. However, too many subintervals can make the calculation more complex and time-consuming.

Can upper and lower sums be used for any type of function?

Upper and lower sums can be used for any function that is continuous over the given interval. However, they are most commonly used for finding the area under a curve or within a region.

What are some common methods for calculating upper and lower sums?

Some common methods for calculating upper and lower sums include the Riemann sum, the trapezoidal rule, and Simpson's rule. Each method has its own advantages and may be better suited for different types of functions.

How do you know if your upper and lower sums are accurate?

To determine the accuracy of your upper and lower sums, you can compare them to the actual area calculated using a more precise method, such as integration. The closer the upper and lower sums are to the actual area, the more accurate they are.

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