Resolving an Integral by Upper and Lower Sums

In summary, the problem is to determine the Upper and Lower Sums that approximate the area under the graph of f(x) = x^3 in the interval [0,1] with a partition of 0,2. The Upper and Lower Sums were calculated and the result was 0.16, but there is concern if the procedure was done correctly.
  • #1
SclayP
27
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So, the problem statement says that i have to determinate the Upper and Lower Sums that aproximate the area under the graph given by the next function: [tex]f(x) = x^3[/tex] in the interval[0,1] with a partition of 0,2

So, i preoceeded to determinate the Upper and Lower Sums but I don't come up with the righ answer (i know because i corroborated by getting the resault of the integral [itex]\int x^3\, dx[/itex] betwen 0 and 1, with my calculator)

P={0; 0,2; 0,4; 0,6; 0,8;1}
[itex]L(f,P) = \sum^{5}_{i=0}[/itex]mi(ti-ti-1) = [itex](0^3)(0,2) + (0,2^3)(0,2) + (0,4^3)(0,2) + (0,6^3)(0,2) + (0,8^3)(0,2) = 0,10[/itex]


That is just plain wrong but i don't know what I am doing wrong...well i won't redact how i did the Upper sums because i guess you got the point...

Thanks.
 
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  • #2
SclayP said:
So, the problem statement says that i have to determinate the Upper and Lower Sums that approximate the area under the graph given by the next function: [tex]f(x) = x^3[/tex] in the interval[0,1] with a partition of 0,2

So, i proceeded to determinate the Upper and Lower Sums but I don't come up with the righ answer (i know because i corroborated by getting the result of the integral [itex]\int x^3\, dx[/itex] between 0 and 1, with my calculator)

P={0; 0,2; 0,4; 0,6; 0,8;1}
[itex]L(f,P) = \sum^{5}_{i=0}[/itex]mi(ti-ti-1) = [itex](0^3)(0,2) + (0,2^3)(0,2) + (0,4^3)(0,2) + (0,6^3)(0,2) + (0,8^3)(0,2) = 0,10[/itex]


That is just plain wrong but i don't know what I am doing wrong...well i won't redact how i did the Upper sums because i guess you got the point...

Thanks.
I get 0.016 --- or as you write it, 0,016 .
 
  • #3
SammyS said:
I get 0.016 --- or as you write it, 0,016 .

Yes, sorry I posted the answer wrong, its 0.16, but what concerns me is that if i did it ok or is wrong the preocedure...
 
  • #4
SclayP said:
Yes, sorry I posted the answer wrong, its 0.16, but what concerns me is that if i did it ok or is wrong the procedure...
The procedure is correct. (Lower sum)
 

FAQ: Resolving an Integral by Upper and Lower Sums

1. What is the purpose of resolving an integral by upper and lower sums?

Resolving an integral by upper and lower sums is a method used to approximate the area under a curve by dividing it into smaller rectangles. This allows for easier calculation of the area and provides a close estimate of the actual integral value.

2. How do you calculate the upper and lower sums for an integral?

To calculate the upper sum, you divide the area under the curve into rectangles with the width of each rectangle being determined by the interval of the integral. The height of each rectangle is determined by taking the maximum value of the function within each interval. The lower sum is calculated in a similar way, but using the minimum value of the function within each interval.

3. Can resolving an integral by upper and lower sums provide an exact solution?

No, resolving an integral by upper and lower sums only provides an approximation of the actual integral value. The more rectangles used, the closer the approximation will be to the actual value, but it will never be an exact solution.

4. What is the significance of using upper and lower sums instead of other methods of approximation?

Using upper and lower sums can provide a more accurate approximation compared to other methods, such as the midpoint rule or the trapezoidal rule. It also allows for a better understanding of the behavior of a function and its integral.

5. Are there any limitations to resolving an integral by upper and lower sums?

Yes, there are limitations to this method. It is only applicable for functions that are continuous and have a defined integral over a given interval. Additionally, the accuracy of the approximation is dependent on the number of rectangles used, so it may not always provide a precise estimation of the integral value.

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