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Numeric Computation of Integrals

Part 5: Increasing the Number of Subintervals

Clearly, when we use more subintervals, our rectangles or trapezoids give better and better approximations to the area under the graph of the function. In the limit as n gets bigger and bigger, the approximations converge to the exact area under the curve.

In this Part, we will see that doubling n gives markedly better improvement in accuracy for some of our approximation methods than it does for others. Throughout this Part, we will be approximating the integral of f(x) = 1/x over the interval [1,3]. You will need the tables of results for this integral you recorded in Parts 1 through 3.

The table below records the errors in the left-hand sum approximation of the integral of f(x)= 1/x over the interval [1,3] for various numbers of approximating rectangles. The error is just the approximate value from the table you calculated in Part 1 minus the exact value of the integral, which is ln(3) = 1.098612289. The third column gives the improvement ratio in the approximation. For example, the improvement for n = 10 over n = 5 is error(5)/error(10) = 0.144989/0.069617 = 2.08.

n for LHS Error Improvement Ratio
5 0.1449885549
10 0.0696167046 2.08
20 0.0340732550
40

  1. Fill in the remaining entries of the above table using your table of results from Part 1. After considering the improvement ratios in column three, what is your conclusion about how much improvement in accuracy to expect when the number of subintervals n is doubled for LHS? What would you expect the error to be when n = 80 subintervals?
  2. Using your table of results from Part 2 where you used trapezoidal sums to approximate the integral of f(x) = 1/x over the interval [1,3], fill in a table in your helper application notebook like the one above, but for TRAP instead of LHS. After considering the improvement ratios in column three of your new table, what is your conclusion about how much improvement in accuracy to expect when the number of subintervals n is doubled for TRAP? What would you expect the error to be when n = 80 subintervals?
  3. Using your table of results from Part 3 where you used midpoint sums to approximate the integral of f(x) = 1/x over the interval [1,3], fill in a table in your helper application notebook like the one above, but for MID instead of LHS. After considering the improvement ratios in column three of your new table, what is your conclusion about how much improvement in accuracy to expect when the number of subintervals n is doubled for MID? What would you expect the error to be when n = 80 subintervals?
  4. Use your Simpson's Rule approximations of the integral of f(x) = 1/x from Part 4 to fill a Simpson's Rule table like the one at the top of this page. After considering the improvement ratios in column three of your new table, what is your conclusion about how much improvement in accuracy to expect when the number of subintervals n is doubled for Simpson's Rule?

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