26.1 Approximating Sums: L n , R n , T n , and M n 811 f x 12345 f x 12345 f x 12 2 34 5 1 1.5 2 2.5 3 3.5 4 4.5 5 f x 12345 f x 12345 OR midpoint rectangles midpoint tangent trapezoids M 8 = 4 5 + 4 7 • 1 2 + 4 9 + 4 11 + 4 13 4 15 4 17 4 19 + + + ( ( = 1.599 L 8 = 1 1 + 2 3 • 1 2 + 2 4 + 2 5 + 2 6 2 7 2 8 2 9 + + + ( ( = 1.828 R 8 = 2 3 + 2 4 • 1 2 + 2 5 + 2 6 + 2 7 2 8 2 9 2 10 + + + ( ( = 1.428 T 8 = L 8 + R 8 = 1.628 We know that 1.428 =R 8 <M 8 =1.599 <  5 1 1 x dx < T 8 = 1.628 <L 8 =1.828 f(x)= 1 x is decreasing and concave up on [1, 5]. Therefore, we know that for any n R n <  5 1 1 x dx < L n and M n <  5 1 1 x dx < T n . If we are interested in more decimal places, we can simply choose larger values of n. 50 Subdivisions Suppose n = 50; we chop [1, 5] into 50 equal pieces each of length x = 5−1 50 = 0.08. We don’t actually want to sum up 50 terms by hand. Work like this is painful to do by hand but it’s child’s play for a programmable calculator or computer. Get out your programmed calculator or computer and check the figures given below. 5 R 50 = 1.577 M 50 = 1.60918 T 50 = 1.60994 L 50 = 1.641 400 Subdivisions Suppose n = 400; we chop [1, 5] into 400 equal pieces each of length x = 5−1 400 = 0.01. We obtain R 400 = 1.60544 M 400 = 1.60943 T 400 = 1.60944 L 400 = 1.61344 Using T 400 as an upper bound and M 400 as a lower bound, we’ve nailed down the value of this integral to 4 decimal places. 5 Yo u’ ll have to enter the following information: the function (often as Y 1 ), the endpoints of integration, and the number of pieces into which you’d like to partition the interval. (And, if you’re using a calculator without L n , R n , T n , and M n programmed, you’ll have to enter the program.) 812 CHAPTER 26 Numerical Methods of Approximating Definite Integrals REMARK M 50 and T 50 give better approximations of  5 1 1 x dx than do R 400 and L 400 . ◆ Summary of the Underlying Principles R n M n T n M n T n R n L n L n If f is increasing on [a, b], then L n <  b a f(t)dt <R n . If f is decreasing on [a, b], then R n <  b a f(t)dt <L n . If f is concave up on [a, b], then M n <  b a f(t)dt <T n . If f is concave down on [a, b], then T n <  b a f(t)dt <M n . For any given n, generally the trapezoidal and midpoint sums are much closer to the actual value of the definite integral than are the left- and right-hand sums. ◆ EXAMPLE 26.2 Approximate  2 0 e −2x 2 dx by finding upper and lower bounds differing by no more than 0.001. 6 SOLUTION This is a great problem on which to practice, because it is impossible to find an antiderivative for f(x)=e −2x 2 unless we resort to an infinite sum of terms. Look at the graph of f(x).It is decreasing on the interval [0, 2] and appears to have a point of inflection somewhere on this interval. 7 f x 1 12 f(x) = e –2x 2 Figure 26.8 Suppose we were planning to use left- and right-hand sums only. As shown in Sec- tion 22.2, the difference between the left- and right-hand sums is given by 6 The function e −kx 2 is of practical importance because, for the appropriate k, its graph gives the bell-shaped normal distribution curve. The area under the normal distribution curve over some given interval is of vital importance to probabilists and statisticians. 7 How can we be sure f is decreasing? f(x)= 1 e 2x 2 .Asxincreases from 0 to 2, 2x 2 increases, so e 2x 2 is positive and increasing. Therefore, its reciprocal is decreasing. Alternatively, d dx e −2x 2 = e −2x 2 (−4x) = −4x e 2x 2 ≤ 0 for 0 ≤ x ≤ 2. 26.1 Approximating Sums: L n , R n , T n , and M n 813 |R n − L n |=|f(b)−f(a)|·x =|f(b)−f(a)|· b−a n . Therefore, in this example |R n − L n |=|e −8 −e 0 |· 2−0 n =|e −8 −1|· 2 n < 1.9994 n . If we want |R n − L n | < 0.001, then we can solve 1.9994 n = 0.001 for n and choose any integer larger than this. n = 1.9994 0.001 = 1999.4, so we can choose n = 2000. If your calculator will accept a number this large, you’re in good shape; the left-hand sum will be an upper bound and the right-hand sum will be a lower bound, because f is de- creasing on [0, 2]. The computation may take some time for the machine to perform, depend- ing upon its power. You should get L 2000 = 0.62711719 and R 2000 = 0.62611753 REMARK Suppose f is monotonic 8 over the interval [a, b], so the actual value of  b a f(x)dx is between L n and R n . Although it is possible simply to try larger and larger values of n until L n and R n are within the desired distance from one another, it is more efficient, if a high degree of accuracy is demanded, to use |R n − L n |=|f(b)−f(a)|x to find an appropriate value of n. Generally, the midpoint and trapezoidal sums give us much better bounds for a partic- ular n.Ifwefind the point of inflection, we can use these sums to form a sandwich around the value of the integral. f(x)=e −2x 2 f  (x) =−4xe −2x 2 f  (x) =−4[e −2x 2 + x(−4xe −2x 2 )] = 4e −2x 2 (−1 + 4x 2 ) f  (x) = 0, where −1 + 4x 2 = 0, that is, where x 2 = 1 4 . On [0, 2] f  (x) = 0atx= 1 2 . Looking back at f  (x) = 4e −2x 2 (−1 + 4x 2 ), we can see that f  changes sign (from negative to positive) at x = 1 2 ,sox=0.5 is a point of inflection. f is concave down on [0, 1 2 ] and concave up on [ 1 2 , 2]. Therefore, 8 f is monotonic on [a, b] if it is either always increasing on [a, b] or is always decreasing on [a, b]. 814 CHAPTER 26 Numerical Methods of Approximating Definite Integrals on [0, 1 2 ], T n gives a lower bound for the integral and M n gives an upper bound; on [ 1 2 , 2], T n gives an upper bound for the integral and M n gives a lower bound. f x 1 1 2 21 f(x) = e –2x 2 Figure 26.9 The combined error on [0, 1 2 ] and [ 1 2 , 2] must be no more than 0.001. We can split up the allowable error however we want. We can play around with the calculator or computer until the sum of |T n − M n | on [0, 1 2 ] and |T m − M m | on [ 1 2 , 2] is less than 0.001. Here is the result of some playing around. If we try n = 50 on both intervals, we obtain the following. On [0, 1 2 ] T 50 = 0.427802 M 50 = 0.4278170 On [ 1 2 ,2] T 50 = 0.198895 M 50 = 0.198759 On each interval the difference between T 50 and M 50 is substantially less than 0.0005. To get the final answer we need to add the lower bounds (T 50 on [0, 1 2 ] and M 50 on [ 1 2 , 2]) to obtain a lower bound for the integral, and add the upper bounds to obtain an upper bound.  2 0 e −2x 2 dx =  .5 0 e −2x 2 dx +  2 .5 e −2x 2 dx lower bound: 0.427802 + 0.198759 = 0.627065 upper bound: 0.428170 + 0.198895 = 0.627065 In the next section we will show an alternative and more efficient method of using the trapezoidal and midpoint sums to approximate this integral. ◆ Numerical methods of integration, such as left- and right-hand sums and trapezoidal and midpoint sums, are very useful not only when we are trying to approximate  b a f(x)dx and can’t find an antiderivative for f , but also in situations in which we don’tevenhavea formula for f . A scientist may be taking periodic data readings, or a surveyor may be taking measurements at preset intervals, and these data sets may constitute the only information we have about f . Numerical methods of integration can be applied directly to the data sets. Exercise 26.1 below deals with information from a data set, and the results of Exercise 26.2 and Exercise 26.3 can be easily applied to data sets in which data have been collected at equally spaced intervals. EXERCISE 26.1 Between noon and 5:00 p.m. water has been leaving a reservoir at an increasing rate. We do not have a rate function at our disposal, but we do have some measurements indicating the rate that water has been leaving at various times. The information is given below. These measurements were not taken at equally spaced time intervals. 26.1 Approximating Sums: L n , R n , T n , and M n 815 Time noon 1:00 1:30 2:00 3:00 3:45 4:15 5:00 Rate out (in gal/hr) 140 160 170 200 250 270 280 300 (a) Find good upper and lower bounds on the amount of water that has left the reservoir between noon and 5:00 p.m. (b) Use a trapezoidal sum to approximate the amount of water that has left the reservoir between noon and 5:00 p.m. Answers are provided at the end of the section. EXERCISE 26.2 Our aim is to approximate  b a f(x)dx when we do not have a formula for f(x).Our data consist of a collection of measurements of f(x) taken at equally spaced intervals. Use a trapezoidal sum to approximate the definite integral. Partition the interval [a, b] into n equal subintervals, each of length x = b−a n . x 0 , x 1 , x 2 , ,are as indicated. x 0 = a and x k = a + kx for k = 1, ,n.y k =f(x k )for k = 0, ,n. Show that the trapezoidal sum can be computed as follows: T n =  1 2  [y 0 + 2y 1 + 2y 3 +···+2y n−1 +y n ]x. x 1 x 0 y 0 y 1 a x 2 x 3 x 4 x n x n–1 y n–1 y n b = = Figure 26.10 The answer is provided at the end of the section. Answers to Selected Exercises Exercise 1 (a) The rate is increasing, so the lower bound is given by the left-hand sum and the upper bound by the right-hand sum. Lower bound: 140(1) + 160(0.5) + 170(0.5) + 200(1) + 250(0.75) + 270(0.5) + 280(0.25) = 897.5 Upper bound: 160(1) + 170(0.5) + 200(0.5) + 250(1) + 270(0.75) + 280(0.5) + 300(0.25) = 1012.5 Between 897.5 and 1012.5 gallons of water have left the reservoir. 816 CHAPTER 26 Numerical Methods of Approximating Definite Integrals (b) The trapezoidal approximation is the average of the left- and right-hand sums given above. The different lengths of the intervals do not change this basic principle. (Con- vince yourself of this.) Trapezoidal sum = 1 2 (897.5 + 1012.5) = 955 Exercise 2. T n = 1 2 [(y 0 + y 1 + y 2 + +y n−1 )x + (y 1 + y 2 + y 3 + +y n )x] = 1 2 (y 0 + 2y 1 + 2y 2 + +2y n−1 +y n )x PROBLEMS FOR SECTION 26.1 Some of the problems in this problem set require the use of a programmable calculator or a computer. These are not highlighted in any way, but common sense should tell you that summing a couple of hundred terms is not a good use of your time. On the other hand, in order to make sure that you understand the numerical methods, some problems explicitly ask you to refrain from using a program to execute the computation. 1. Suppose we use right- and left-hand sums to approximate  b a f(t)dt.Wepartition the interval [a, b] into n equal pieces each of length t. Let R n be the right-hand sum using n subdivisions and L n be the left-hand sum using n subdivisions. (a) Show that R n = L n + f (b)t − f (a)t. (b) Conclude that R n − L n = [f(b)−f(a)]t = [f(b)−f(a)] |b−a| n . 2. (a) Find  x 1 1 t dt, where x>0. (b) ln x can be defined to be  x 1 1 t dt. Sketch the graph of 1/t and approximate ln 2 numerically by partitioning the interval [1, 2] into 4 equal pieces and computing L 4 and R 4 . Write out the sum long-hand, without using a calculator program. (c) Into how many equal pieces would you have to subdivide the interval [1, 2] in order to approximate ln 2 so that R n and L n give upper and lower bounds that differ by no more than 0.01? Call this number p. (d) Compute M p and T p . (e) Into how many equal pieces would you have to subdivide the interval [1, 3] in order to approximate ln 3 so that R n and L n give upper and lower bounds that differ by no more than 0.001? That differ by no more than D? 3. Compute  e 1 ln xdxwith error less than 0.002. Try, by experimentation, to see how many subdivisions are required for M n and T n to differ by less than 0.002. How many are required for L n and R n to differ by less than 0.002? 26.1 Approximating Sums: L n , R n , T n , and M n 817 4. (a) Let f be the function graphed below. f is increasing and concave down. f x a b Let A =  b a f(x)dx. Suppose that estimates of f(x)dx are computed using the left, right, and trapezoid rules, each with the same number of subintervals. We’ll denote these estimates by L n , R n , and T n , respectively. Put the numbers A, L n , R n , and T n in ascending order. Justify your answer and explain why your answer is independent of the value of n used. (b) Answer the same question as above if the graph of f is the one given below. f is decreasing and concave up on [a, b]. f x a b 5. Consider  4 1 √ xdx. (a) Find a value of n for which    L n −  b a f(x)dx    ≤0.01. (b) Use the value of n from part (a) to find L n and R n . (c) Is the average of the left- and right-hand sums larger than the integral, or smaller? (d) Compare your numerical approximations to the answer you get using the Funda- mental Theorem of Calculus. 6. Approximate  1 0 √ 1 + x 4 dx with error less than 0.01. 7. Give upper and lower bounds for  2 0 10 2+x 5 dx such that the upper and lower bounds differ by less than 0.01. 8. Give upper and lower bounds for  3 2 1 ln x dx such that the two bounds differ from one another by less than 0.05. Explain how you know that the upper bound is indeed an upper bound and the lower bound is indeed a lower bound. 9. Consider  9 4 1 √ x dx. (a) Find a value of n for which | L n − R n | ≤ 0.01. (b) Use that value of n to find L n and R n . 818 CHAPTER 26 Numerical Methods of Approximating Definite Integrals (c) If you take the average of the left- and right-hand sums, will your approximation be larger than the integral, or smaller? (d) Compare your numerical approximations to the answers you get using the Funda- mental Theorem of Calculus. 10. One of your friends is doing his mathematics homework on the run. He managed to squeeze in one problem between lunch and his expository writing class. He used his calculator to find L n , R n , T n , and M n to approximate a definite integral and jotted down the results on a napkin. They were 0.367617, 0.3211885, 0.341189, and 0.274760. He didn’t write down the problem, nor did he record the value of n he used. But he did sketch the function (over the relevant interval) on the napkin; it looked like this. f x He wants your help in labeling the data he recorded as a left-hand sum, a right-hand sum, a midpoint sum, and a trapezoidal sum. Which is which? Explain your answer. 11. Measurements of the width of a pond are taken every 20 yards along its length. The measurements are: 0 yards, 60 yards, 50 yards, 70 yards, 50 yards, and 30 yards. Approximate the surface area of the pond using the Trapezoidal rule. 60 50 70 50 30 20 20 20 20 20 12. Approximate each of the following integrals with error less than 1/100. (Note: If you look at all the questions and think about a strategy in advance, you will only have to compute two integrals in order to answer all four questions with the desired degree of accuracy. There is no problem with being more accurate than is requested.) Briefly explain what you have done and how many subdivisions you used. (i)  1 0 e −( 1 2 )x 2 dx (ii)  2 1 e −( 1 2 )x 2 dx (iii)  2 0 e −( 1 2 )x 2 dx (iv)  2 −2 e −( 1 2 )x 2 dx 26.1 Approximating Sums: L n , R n , T n , and M n 819 13. Let f(x)= 3 x 3 +x . Let a and b be positive constants, 0 <a<b.Below are two partitions of the interval [a, b]. One (given with w’s) partitions [a, b] into 8 equal subintervals, each of length w; the other (given with t’s) partitions the interval into 12 equal pieces each of length t. w i = a + iw, i = 0, 1, ,8 t i =a+it, i = 0, 1, ,12 Put the following expressions in ascending order, with “<” or “=” signs between them. (Identify the sum with the letter preceding it.) A = 7  i=0 f(w i )w B = 8  i=1 f(w i )w C = 11  i=0 f(t i )t D = 12  i=1 f(t i )t E =  b a f(w)dw F =f(a)·(b − a) 14. Approximate  2 −1 arctan xdxusing left- and right-hand sums to obtain an upper and lower bound for the integral with difference less than 0.05. Save time by graphing y = arctan x and using symmetry to simplify the problem. 15. The function f(x)is decreasing and concave down on the interval [3, 5]. Suppose that you use a right-hand sum, R 100 , a left-hand sum, L 100 , a trapezoidal sum, T 100 , and a midpoint sum, M 100 , all with 100 subdivisions, to estimate  5 3 f(x)dx. Select all of the following that must be true. (a) L 100 ≥ R 100 (b)  5 3 f(x)dx ≥T 100 (c) T 100 ≥ R 100 (d) T 100 ≥ M 100 (e) M 100 ≥ L 100 (f) T 100 = (L 100 + R 100 )/2 16. Suppose that g is a differentiable function whose derivative is g  (x) = 2 x 2 +3 . Partition [0, 2] into n equal pieces each of length x and let x k = kx, where k = 0, 1, ,n.Put the following expressions in ascending order (with “<” or “=” signs between them). A = n  i=1 g(x i )x B = n−1  i=0 g(x i )x C = lim n→∞ n−1  i=0 g(x i )x D = lim n→∞ n  i=1 g(x i )x 820 CHAPTER 26 Numerical Methods of Approximating Definite Integrals 17. Suppose L 10 <  b a f(x)dx <R 10 and |R 10 − L 10 | < 0.01. Which of the following statements must be true? Circle all such statements. (a) |R 10 −  b a f(x)dx|<0.01 (b) |T 10 −  b a f(x)dx|<0.01 (c) |T 10 −  b a f(x)dx|<0.005 (d) |L 10 −  b a f(x)dx|<0.005 18. Two of the following three integrals you can evaluate exactly. One you cannot, until learning integration by parts. Identify the one you cannot evaluate exactly and approx- imate it with an error under 0.05. Find exact answers for the other two integrals. (a)  e 1 ln x x dx (b)  1 0 xe x 2 dx (c)  1 0 xe x dx 26.2 SIMPSON’S RULE AND ERROR ESTIMATES Simpson’s Rule We began our discussion with left- and right-hand sums, sums that provide bounds for  b a f(x)dx if f is monotonic on [a, b]. Then we came up with the clever idea of using the average of L n and R n to approximate a definite integral. This led us to the trapezoidal sum, which generally gives a much better approximation to a definite integral than either the left or right sums for the same number of subdivisions. Encouraged by our success, we might start thinking about averaging the midpoint and trapezoidal sums. An even better idea would be to take a weighted average of these two sums. This is because the midpoint rule generally gives us a better fit than does the trapezoidal rule. In fact, analysis of numerical data indicates that the error in using the trapezoidal rule is about two times the size as that from using the midpoint rule. From the picture below this seems plausible. The weighting M n +M n +T n 3 = 2M n +T n 3 gives us a method of approximating definite integrals known as Simpson’s rule. x i x i +1 midpoint = = Error using trapezoid = Error using midpoint tangent trapezoid The shaded area is about double the blackened area. Weighted average: weight midpoint twice as heavily as the trapezoidal sum. 2M n + T n 3 Simpson's rule Figure 26.11 Simpson’s rule: S 2n = 2M n + T n 3 . show an alternative and more efficient method of using the trapezoidal and midpoint sums to approximate this integral. ◆ Numerical methods of integration, such as left- and right-hand sums and. and left-hand sums to approximate  b a f(t)dt.Wepartition the interval [a, b] into n equal pieces each of length t. Let R n be the right-hand sum using n subdivisions and L n be the left-hand. Approximate  2 −1 arctan xdxusing left- and right-hand sums to obtain an upper and lower bound for the integral with difference less than 0.05. Save time by graphing y = arctan x and using symmetry to simplify