#$&*
Mth 173
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
Use both 2-interval and a 5-interval approximations to find the left and right Riemann sums of the function y = 1.1 * cos( .18 t) from t = 1.92 to t = 6.08.
• Based on these approximations make your best estimate of the integral.
• What would be the difference between left- and right-hand sums if we used 100 intervals for the approximation?
** **
This is a problem a ran into while trying to study for your test. I can not figure out how to solve it. I look back through the notes for Riemann Sums but the only notes there are involve using some type of syntex that I am unsure of how to optain. So could you please show me how to work this problem?
** **
Could you please show me how to work this problem?
@& Recall the trapezoidal approximation, where we broke a larger interval into subintervals, each of which gave us a trapezoid. We approximated the area beneath the graph of the curve by finding the areas of the trapezoids and adding them up. The average altitutde of each trapezoid was the average of the two function values; the width of each was that of the corresponding subinterval.
In the case of left and right Riemann sums, we just approximate the area using rectangles. There is a rectangle for each subinterval, whose width is equal to that of the subinterval.
We find the altitude as follows:
For the left Riemann sum the altitude used for the subinterval is the value of the function at the left-hand endpoint of the subinterval.
For the right Riemann sum the altitude used for the subinterval is the value of the function at the right-hand endpoint of the subinterval.
A midpoint approximation would use as the altitude the value of the function at the midpoint of the interval.
A trapezoidal approximation, mentioned above, uses as the altitude the average of the function values at the left and right-hand endpoints of the interval.
To do a 2-interval approximation for the function you are given, first divide the interval 1.92 <= t <= 6.08 into two subintervals. The two subintervals would be 1.92 <= t <= 4 and 4 <= t <= 6.08.
Evaluate the function at t = 1.92, the left-hand endpoint of the first subinterval.
Evaluate the function at t = 4, the right-hand endpoint of the first interval (and the left-hand endpoint of the second).
Evaluate the function at t = 6.08, the right-hand endpoint of the second subinterval.
Using the function values at the left-hand endpoints, and the widths of the intervals (both intervals are of width 2.08), find the areas of the corresponding rectangles. Add up the areas. This is your left-hand approximation.
Do the same, but use the right-hand endpoints of the intervals.
Based on your results make your best estimate of the integral.
Then repeat the process for five subintervals. The interval 1.92 <= t <= 6.08 has width 4.16, so the subintervals will have length 4.16 / 5 = .832. The first, for example, will run from 1.92 to 1.92 + .832 = 2.752. Evaluate the function at 1.92 and 2.752 to get the left- and right-hand values for this interval. Then move on to the second interval, from 2.752 to 2.752 + .832, etc..
*@