Calculating Riemann Sums

Calculating Riemann Sums Overview The Riemann integral is defined as a limit of Riemann sums (see Sections 5.2 and 5.3 of the textbook). The limiting process in the definition of the Riemann integral is complicated. It is impractical, if not impossible, to evaluate most Riemann integrals with pencil and paper from the definition. However, it is possible, and even practical, to approximate most Riemann integrals very accurately from the definition by using a calculator. In this lab you will learn to use your calculator to calculate Riemann sums efficiently. In the next lab you will use these calculator techniques to investigate how the lengths of the subintervals and the selection of the evaluation points used in the Riemann sum affect the accuracy with which a Riemann sum approximates an integral. Recall that ∫ 2 −1 n ∑ x 3 + 1 dx means the limit of the Riemann sums Sn = k =1 ck 3 + 1 ∆x . In these sums, n is the number of subintervals into which the interval from –1 to 2 is divided by equally spaced partition points 2 − (−1) 3 –1 = x0 < x1 < … < xn-1 < xn = 2. The length of each subinterval is Δx = = . (The subintervals don’t n n have to be all the same length, but calculations are easier when they are.) Each ck is a number from the kth subinterval. This limit statement tells us that ∫ 2 −1 x 3 + 1 dx is the number that can be approximated as closely as anyone could want (though not necessarily exactly) by Riemann sums n ∑ ck 3 + 1 ∆x computed using a k =1 partitioning of the interval – 1 ≤ x ≤ 2 into a large enough number of equal subintervals. The partition points x0 < x1 < x2 < … < xn-1 < xn that divide the interval of integration – 1 ≤ x ≤ 2 into subintervals are at the heart of the idea of a Riemann sum. But the evaluation points c1, c2, c3, c4, … , cn-1, cn are more prominent in calculating the value of a Riemann sum. The key to learning to use your calculator to evaluate Riemann sums is to learn to calculate with LISTS of evaluation points. I.1. We begin by calculating several Riemann sums S6 for the integral ∫ 2 −1 x 3 + 1 dx based on dividing the interval of integration into 6 subintervals of equal length. On the graph of f( x) = x3 + 1 (next page), divide the interval [ -1, 2] on the x-axis into 6 equal subintervals. Mark and label the endpoints of these subintervals x0, x1, x2, x3, x4, x5, and x6 . The seven endpoints of these six subintervals are the partition points. We’ll first calculate a Riemann sum using the left endpoints of the subintervals as the evaluation points. With this choice the evaluation points c1, c2, c3, c4, c5, c6 coincide with the first six partition points x0, x1, x2, x3, x4, x5. Step 1: Read your calculator manual to find out how to create a list. Enter the list { – 1, – 0.5, 0, 0.5, 1, 1.5 } of left endpoints of the subintervals from above into your calculator. Store this list under a name that suggests Left Endpoints. (I suggest LFTND or lftnd.) Step 2: To compute a Riemann sum S6 we need to evaluate the integrand f( x) = x3 + 1 at each evaluation point in the list LLFTND. TI® calculators will create a new list by evaluating a function at each entry of a specified list. Enter the integrand f(x) = x3 + 1 as Y1 on the Y= screen of your TI-83® or TI-84® (or as y1 on the GRAPH screen of your TI-86® or TI89®). Return to the home screen and enter Y1(LLFTND). On a TI-83® or TI-84®, to enter LLFTND, press [2nd] [LIST], scroll to LLFTND and press [ENTER]. On a TI-89®, to enter lftnd, press [2nd] [VAR-LINK], scroll to lftnd and press [ENTER]. Press [ENTER] a second time. A list of the values of the integrand at the six evaluation points in the list LLFTND will be displayed. You may want to store this list under a name that suggests Function Values. (I suggest FNVLS or fnvls.) The numbers in this list are the successive heights of the 6 rectangles that approximate the area under the graph and, hence, the integral. Draw these approximating rectangles on the graph on the next page. Step 3: To get the areas of the approximating rectangles, we must multiply each of the function values in the list LFNVLS (or fnvls) you just created by the length of the bases of the approximating rectangles. The length of 2 − (−1) the base of each rectangle is Δx = = 0.5. Multiplying a list by a number multiplies each entry of the 6 list by that number. To find the area of each of the approximating rectangles in the Riemann sum, multiply 0.5 times the list of function values. (On a TI-83®, TI-84®, or TI-89® enter 0.5* [2nd ], (ANS) and press [ENTER].) Step 4: The value of the Riemann sum S6 calculated with left endpoints as evaluation points is the sum of the six entries in the list of areas you just created. The SUM command adds all the entries in a list. (To find the SUM command on a TI-83® or TI-84®, press [2nd ], (LIST), scroll to MATH, and choose SUM from the menu. On a TI-89®, press [2nd ], (MATH), scroll to LIST, press the right arrow, and choose SUM from the menu.) Apply the SUM command to the list of areas of rectangles (use sum(2nd ANS) ) and verify that S 6 ( left endpoint evaluations) = 3.250969074 (from a TI-84®). Page total ________ I.2. The approximation for ∫ 2 −1 (a) Use the graph of f(x) = x 3 + 1 dx calculated in I.1 is not very accurate. x3 + 1 above to explain why S6 calculated with left endpoint evaluations is smaller than the actual value of ∫ 2 x 3 + 1 dx . −1 (b) Without evaluating the integral or finding the actual area, estimate the amount (not the percentage) by which S6 calculated with left endpoint evaluations differs from the actual value of ∫ 2 −1 x 3 + 1 dx . Explain how you arrived at this estimate. (Method for estimating error must be plausible. Explanation must be consistent with method used to estimate the error.) (c) Riemann sum approximations for ∫ 2 −1 x 3 + 1 dx calculated using left endpoints as evaluation points will a lway s be smaller than the actual value of the integral. What property of the function accounts for this? Use a graph and words to explain why it is that if a function y = f( x) has the property you just stated, then e v er y Riemann sum approximation for ∫ 2 f ( x) dx calculated using left endpoints as evaluation points −1 will be smaller than the actual value of the integral. Page total ________ II.1. Next, calculate the Riemann sum S6 for the integral ∫ 2 −1 x 3 + 1 dx using the right endpoints of the subintervals as evaluation points. To get the right endpoints of the subintervals, add the length of the subintervals (Δx = 0.5) to the left endpoints. To recall the list of left endpoints { -1, -0.5, 0, 0.5, 1, 1.5 } from memory press [2nd], (LIST), scroll to the name you gave the list of evaluation points, and press [ENTER]. Adding a number to a list adds the number to each entry of the list, so just add 0.5 to this list. (Enter 0.5+ [2nd ] (ANS), [ENTER].) Store this new list under a name that suggests Right Endpoints. (I suggest RTND or rtnd.) Follow Steps 2 - 4 from #1 to compute S6 using right endpoint evaluations. Illustrate (label!) and interpret (explain!) all steps of this calculation on the graph of f(x) = x3 + 1 below. S 6 (right endpoint evaluations) = _______________________________ (to full accuracy of your calculator) Page total ________ II.2 The approximation for ∫ 2 −1 (a) Use the graph of f(x) = x 3 + 1 dx using right endpoint evaluations is not very accurate. x3 + 1 above to explain why S6 calculated with right endpoint evaluations is larger than the actual value of ∫ 2 −1 x 3 + 1 dx . (b) Without evaluating the integral or finding the actual area, estimate the amount (not the percentage) by which S6 calculated with right endpoint evaluations differs from ∫ 2 −1 x 3 + 1 dx . Explain how you arrived at this estimate. (c) Riemann sum approximations for ∫ 2 −1 x 3 + 1 dx calculated using right endpoints as evaluation points will a lway s be larger than the actual value of the integral. What property of the function accounts for this? Use a graph and words to explain why it is that if a function y = f( x) has the property you just stated, then every Riemann sum approximation for ∫ 2 f ( x) dx calculated using right endpoints as evaluation points −1 will be larger than the actual value of the integral. Page total ________ III.1. Calculate S6 using mid-point evaluations. Illustrate and interpret all steps of the calculations on the graph of f(x) = x3 + 1 below. S 6 (m idpoint evaluations) = _______________________________ (to full accuracy of your calculator) III.2. Do you think the approximation to the integral calculated using midpoint evaluations is larger or smaller than the actual value of the integral? Explain without finding or using the exact value of the integral why you think so. Page total ________ IV.1. (a) Most calculators will evaluate integrals numerically. Use the manual for your calculator to learn how to evaluate integrals numerically. Evaluate ∫ 2 −1 x 3 + 1 dx and record the result. __________________ (b) Which of the approximations to the integral calculated above is closest to the value your calculator gives for the integral? Explain why one might expect that. (c) If possible, draw the graph of a continuous function y = f(x ) on the interval [ –1, 2] and a partition of the interval [ –1, 2] so that the Riemann sum calculated using left endpoint as evaluation points gives a better approximation for ∫ 2 f ( x) dx than the Riemann sum calculated using midpoints as evaluation −1 points. If it is not possible to graph such a function and partition, explain how you know. Page total ________