Approximating integral using riemann sums

In summary, the conversation discusses finding lower and upper Riemann sums and using them to approximate the definite integral. The most accurate approximation would be to average the right and left-hand sums and the error can be found by taking the absolute value of the difference between the two sums. It is important to ensure that the heights and widths of the rectangles are chosen correctly for accurate results.
  • #1
missavvy
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0

Homework Statement


f: [0,1] -> Reals, f(x) = 3-x2
P={0,1/2,1}

Find lower and upper Riemann sums, and approximate the definate integral using them and find the corresponding approximation error.


Homework Equations





The Attempt at a Solution


So I tried finding the upper Riemann sum
Spf = 3(1/4) + 2(1/2) = 19/8
and the lower:
spf = 3(1/2) + (3-1/4)(1/2) = 23/8

Is this correct? If so, how do I approximate the definate integral? Do I just pick the upper or lower sum and use that as my approximation? And how would I find the error?

Any help is appreciated :)
 
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  • #2
The most accurate approximation would be to average the RHS and LHS. The error would be the absolute value of the difference between the RHS and LHS, or another way of looking at it is that its the absolute value of (f(b)-f(a))Δt, either way they should be non-negative.

As for the sums themselves, they way you wrote it a little confusing, but all you're doing is creating rectangles either above or below f(x) depending on the direction of the sum, that is left or right, and whether it increasing or decreasing.

Hope this helps a bit.
 
  • #3
Ok.. is Δt the difference between the x_j and x_j-1?

I added the upper and lower sums and divided by 2 and got 2.625 which is really close to the actual value of 2.6666...7.
For the error, I took the absolute value:
|19/8-23/8| = 0.5
So then 0.5 is the approximation error.. ?
 
  • #4
hi, I just thought I should point out that your upper riemann sum is less than your lower riemann sum.

I'm sure you're well aware that for any parition P, we have that Spf >= spf.

I suggest redrawing your graph making sure that you have chosen the appropriate height/width for the rectangles. good luck
 

Related to Approximating integral using riemann sums

1. What is a Riemann sum?

A Riemann sum is a method for approximating the area under a curve by dividing the region into smaller rectangles and calculating the sum of their areas.

2. How is a Riemann sum used to approximate integrals?

A Riemann sum is used by dividing the interval on which the integral is to be evaluated into equally spaced subintervals and finding the sum of the areas of the rectangles formed by the height of the function at each subinterval.

3. What is the difference between a left, right, and midpoint Riemann sum?

In a left Riemann sum, the height of each rectangle is determined by the function's value at the left endpoint of the subinterval. In a right Riemann sum, the height is determined by the function's value at the right endpoint. In a midpoint Riemann sum, the height is determined by the function's value at the midpoint of the subinterval.

4. How do you choose the number of subintervals for a Riemann sum?

The number of subintervals, or rectangles, used in a Riemann sum can vary depending on the desired level of precision. Generally, the more rectangles used, the more accurate the approximation will be. However, too many rectangles can become computationally intensive, so a balance must be struck.

5. Can a Riemann sum give an exact value for an integral?

No, a Riemann sum can only approximate the value of an integral. The more rectangles used, the closer the approximation will be to the exact value, but it will never be exact. To find the exact value, the integral must be evaluated using other methods such as the Fundamental Theorem of Calculus.

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