6.5 Numerical Approximations of Definite Integrals

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1 Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 6.5 Numericl Approximtions of Definite Integrls Sometimes the integrl of function cnnot be expressed with elementry functions, i.e., polynomil, trigonometric, exponentil, logrithmic, or suitble combintion of these. For exmple, the functions e x, sin (x ) nd sin x x don t hve simple ntiderivtives, i.e., it is not possible to write down simple nlytic formul in terms of elementry functions. In cses like these we still cn find n pproximte vlue for the integrl of such functions on finite intervl. Left-endpoint nd Right-endpoint Approximtions We lredy know tht we cn pproximte definite integrl f(x)dx by left-endpoint Riemnn sum or right-endpoint Riemnn sum for some finite n : L n f(x i 1 ) x nd R n f(x i ) x where x 0, x 1,, x n re n + 1 eqully spced points in [, b] with x 0, x n b nd x i + i x, 0 i n where x b n. Exmple Approximte xdx using left-endpoint Riemnn sum nd then rightendpoint Riemnn sum with 4 intervls, i.e., n 4. Solution. We first construct the following chrt. Thus, x i f(x i ) L 4 (f(1) + f() + f(3) + f(4)) x ( )

2 nd R 4 (f() + f(3) + f(4) + f(5)) x ( ) For comprison, we know tht the exct vlue of the integrl we re seeking to pproximte is 5 1 dx ln x 1 Thus, for n 4 the left-endpoint nd right-endpoint Riemnn sums give poor pproximtions. Figures 6.5.1() nd 6.5.1(b) illustrte why the left nd right rules re so inccurte Figure By drwing pictures of the geometric regions involved one cn esily see tht nd L n f(x)dx R n, when f(x) is incresing R n f(x)dx L n, when f(x) is decresing. So if f(x) is incresing then L n is n underestimte while R n is n overestimte. Similrly, when f is decresing R n is n underestimte nd L n is n overestimte.

3 Error in Left Riemnn nd Right Riemnn Sums With bit of work, it cn be shown tht nd f(x)dx L n f(x)dx R n K(b ) n K(b ) n where f (x) K on the intervl [, b]. Thus, doubling the number of intervls will decrese the error by fctor of 1. Also, since f (x) K on [, b], the error depends on how steeply the grph of f rises or flls. See Figure Figure 6.5. Exmple 6.5. Let A x dx Then by the definition of L nnd R n we hve L n f(x i 1 ) x 1 9 x i 1 n n (i 1) n 9 n + 9(i 1). 3

4 Using TI-83, one cn find L n for ny n. For exmple, L 160 cn be found s follows: Hit nd List (the stt button). Go to MATH nd select sum( nd press enter. You will see prompt sum(. Go bck to nd List. Now go to OPS nd select seq( nd press enter. Type nd then hit enter. Similrly, sum(seq(9/( (N 1)), N, 1, 160, 1)) R n Thus, we hve the following tble 9 n + 9i. n R n L n A R n A L n In greement with the result bove, the errors in the right-hnd rule or lefthnd rule pproximtions decrese by fctor of, roughly, 1 when we double the number of intervls. Also, note tht the left errors re ll negtive since the function 1 x is decresing so tht L n is n overestimte. Similr remrk for the right errors Midpoint Approximtion Insted of using the left- or right-endpoints we use the midpoint. Recll tht the midpoint of n intervl [x, y] is given by the midpoint formul x+y. The midpoint pproximtion is given by M n ( ) xi 1 + x i f x. Exmple Approximte xdx with the midpoint rule using 4 intervls. Solution. Let m i denote the midpoint of the intervl [x i 1, x i ]. Then 4

5 Thus, m i f(m i ) M 4 (f(1.5) + f(.5) + f(3.5) + f(4.5)) x This nswer is firly close to the exct nswer. The midpoint rule pproximtes with rectngles on ech subdivision tht re prtly bove nd prtly below the grph, so the errors tend to blnce out. See Figure Figure Note tht if the grph of f(x) is concve down on [, b] then M n is n overestimte of f(x)dx since the midpoint rectngle nd the trpezoid constructed by drwing the tngent line to the grph t the midpoint hve the sme re since the tringles ABC nd CEF re equl. (See Figure 6.5.4) Figure

6 On the other hnd, if f(x) is concve up on [, b] then M n is n underestimte of f(x)dx. See Figure Figure Error in Midpoint Rule An nlysis of error similr to ones discussed before shows tht K(b )3 f(x)dx M n 4n, where f (x) K in [, b]. Hence, doubling the number of intervls will decrese the error by, roughly, fctor of 1 4. Exmple Let A x dx. Then by the definition of M n we hve M n f(m i ) x 1 m i 9 n Using TI-83 we hve the following tble 9 x i 1 + x i n 18 n + 18i 9. 6

7 n M n A M n Notice tht errors in the midpoint rule pproximtions decrese by fctor of, roughly, 1 4 when we double the number of intervls. Note tht the errors re ll positive since the function 1 x is concve up so tht the midpoint rule is n underestimte. Note lso, tht the error in ech pproximtion is pproximtely 1 of the corresponding error for the trpezoid rule Trpezoid Rule There is no reson why we should necessrily use rectngles to pproximte b f(x)dx. We could use the function vlues of both endpoints of the intervl nd pproximte the intervl by trpezoids insted. Recll tht the re of trpezoid with bse [x i 1, x i ] nd sides f(x i 1 ) nd f(x i ) is given by f(x i 1 ) + f(x i ) x. Thus, the trpezoid pproximtion is given by f(x i 1 ) + f(x i ) T n x L n + R n. Exmple Approximte xdx with the trpezoid rule using 4 intervls. Solution. Using the following chrt we cn write x i f(x i ) f(1) + f() f() + f(3) f(3) + f(4) T 4 ( f(4) + f(5) ) x

8 This nswer is firly close to the exct nswer Note tht if the grph of f(x) is concve up then the re of ech trpezoid is lrger thn the re under the grph so tht T n is n overestimte of b f(x)dx. See Figure 6.5.6(). On the other hnd, if the grph of f(x) is concve down then the re fo ech trpezoid is smller thn the re under the grph so tht T n is n underestimte of the definite integrl. See Figure 6.5.6(b). Figure It follows from the bove discussions tht: (i) If f(x) is concve up nd incresing on [, b] then L n M n f(x)dx T n R n. (ii) If f(x) is concve up nd decresing on [, b] then R n M n f(x)dx T n L n. (iii) If f(x) is concve down nd incresing on [, b] then L n T n f(x)dx M n R n. (iv) If f(x) is concve down nd decresing on [, b] then R n T n f(x)dx M n L n. 8

9 Error in the Trpezoid Rule It cn be shown tht the bsolute vlue of the error in the trpezoid rule is bounded by K(b )3 f(x)dx T n 1n where f (x) K in the intervl [, b]. Thus, if we double the number of intervls then we should expect the error to decrese by fctor of 1 4. Exmple Let A x dx. Then by the definition of T n we hve f(x i 1 ) + f(x i ) T n x ( ) 9 x i 1 x i n ( 1 n n + 9(i 1) + Using TI-83 we hve the tble n n + 9i n T n A T n ) 9 n. We see tht the errors in the trpezoid rule pproximtions re significntly smller thn the corresponding errors for the left-hnd nd right-hnd rule pproximtions. Moreover, in greement with the result bove, the errors in the trpezoid rule pproximtions decrese by fctor of, roughly, 1 4 when we double the number of intervls. Note tht the errors re ll negtive due to the fct tht the function 1 x is concve up so the trpezoid rule is n overestimte. Note lso, tht the trpezoid rule converges to the vlue of the definite integrl t significntly fster rte thn do the left-hnd rule or the right-hnd rule. Also, the error of the midpoint rule is 1 the size of the error of the trpezoid rule 9

10 Remrk The errors in either the midpoint rule or the trpezoid rule not only depend on n but lso on the size of f nd hence on the concvity of f, i.e. on how bent the curve is. See Figure Figure Simpson s Approximtion The trpezoid rule discussed in the previous section uses line segments to pproximte the grph of the integrnd, i.e. we pproximte the re under the grph by using trpezoids. In this section, we will introduce n pproximtion technique, known s Simpson s rule, tht pproximtes the grph of the integrnd by using prbols insted (i.e. functions with equtions y Ax + Bx + C). The process strts by dividing the intervl [, b] into n equl subintervls ech of length x b n using the prtition points x 0 < x 1 < x < < x n b, where x i + i x, 0 i n. On the intervl [x i 1, x i ] we wnt to pproximte f(x) by qudrtic function,i.e. such tht f(x) Ax + Bx + C f(x i 1 ) Ax i 1 + Bx i 1 + C f(x i ) Ax i + Bx i + C f(m i ) Am i + Bm i + C 10

11 where m i is the midpoint of [x i 1, x i ]. Thus, xi xi f(x)dx (Ax + Bx + C)dx x i 1 x i 1 [ A 3 x3 + B ] xi x + Cx x i 1 A 3 (x3 i x 3 i 1) + B (x i x i 1) + C(x i x i 1 ) A 3 (x i x i 1 )(x i + x i x i 1 + x i 1) + B (x i x i 1 )(x i + x i 1 ) + C(x i x i 1 ) [ A(x i + x i x i 1 + x i 1) + 3 B(x i + x i 1 ) + 3C x 3 But ( ) f(x i 1 ) + 4f(m i ) + f(x i ) Ax xi 1 + x i i 1 + Bx i 1 + C + 4A ( ) xi 1 + x i +4B + 4C + Ax i + Bx i + C [ A(x i + x i x i 1 + x i 1) + 3 ] B(x i 1 + x i ) + 3C. It follows tht xi Hence, f(x)dx x x i 1 3 f(x)dx 1 3 ( f(xi 1 ) + f(m i ) + f(x ) i). ( ) f(xi 1 ) + f(x i ) x + f(m i ) x M n + T n. 3 We denote the expression on the right-hnd side by S n. Exmple Use Simpson s rule to pproximte the vlue of π. ]. 11

12 Solution. First recll tht dx rctn x 1 1+x 0 π 4. Let n in Simpson s rule so tht x 1, x 0 0, x 1 1, x 1, m 1 1 4, nd m 3 4. Thus, x dx S x ( f(x0 ) + f(m 1 ) + f(x 1) ( ) f(x 1) + f(m ) + f(x ) ) This produces the pproximtion π dx x Error of Simpson s Rule It my be shown tht the bsolute vlue of the error using Simpson s rule is f(x)dx S n K(b )5 90n 4, where f (4) (x) K in [, b], resulting in drmtic improvement over both the trpezoid nd midpoint rules. For Simpson s rule, doubling the number of intervls decreses the error by, roughly Exmple Let A xdx. Then by the definition of the Simpson s rule we hve S n M n + T n. 3 Using Exmples nd we hve the following tble n S n A S n