Chapter7 - 7.1 l The Substitution Rule 333 Figure 7.6 The...

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Unformatted text preview: 7.1 l The Substitution Rule 333 Figure 7.6 The region corresponding to the definite integral in Example 10. Therefore, 9 2 62 5 f dx=f —du=21n|ul]1=2(ln6—ln1)=21n6 I 4 x‘3 i H. We can easily spend a great deal of time on integration techniques. The problems can get very involved, and to solve them all, we would need a big bag full of tricks. There are excellent software programs (such as Mathematich and MATLAB®) that can integrate symbolically. These programs do not render integration techniques useless; in fact, they use them. Understanding the basic techniques conceptually and being able to apply them in simple situations makes such software packages less of a “black box.” Nevertheless, their availability has made it less important to acquire a large number of tricks. So far, we have learned only one technique: substitution. Unless you can immediately recognize an antiderivative, substitution is the only method you can try at this point. As we proceed, you will learn other techniques. An additional complication will then be to recognize which technique to use. If you don’t see right away what to do, just try something. Don’t always expect the first attempt to succeed. With practice, you will see much more quickly whether or not your approach will succeed. If your attempt does not seem to work, try to determine the reason. That way, failed attempts can be quite useful for gaining experience in integration. 3 Section 7.1 Problems I 7.1.1 In Problems 1—16, evaluate the indefinite integral by making the given substitution. 1. flu/Firinwithu =xz+3 2. fsxVPTidx,wiihu =x3+l 3. [am —x2)‘/“dx, with u =1 —.r2 4. /4x3(4+x‘)‘/3dx, withu = 4 +x‘ 5. f5cos(3x)dx. with u = 3x 6. f5sin(1- 2.x)dx, withu = 1 —2x 7. /7xzsin(4x3)dx, with u = 4x3 8. fxcos(x2 —- 1)dx, with u = x2 —1 9. fem’dx, with u = 2x +3 10. f3e"‘ dx,with u =1- x 11. / xe-xz/de, with u = —x2/2 12. [xel‘l‘2 dx, with u = 1 — 3x2 .s...;-i.-. x Av:r.-a A-«r .»«e--.—'—.—.: 7v: W W . i ii iii [Tl 334 Chapter7 El Integration Techniques 13.] "+2 dx,withu=x2+4x x2 + 4x 14. [Sixxzdx.withu=3—x 15.] 3x4dx,withu=x+4 x+ 16.] " dx,withu=5—x 5—): 2 In Problems 17—36, use substitution to evaluate the indefinite integrals. 17. [Wt + 3dx 19. [(4x —3) 2x2 -3x +2dx 13. f (4 — x)”’ dx 20. f (x2 -— 2x)(x3 — 3x2 + 3)"3 dx 31. f tanx sec2 x dx 2 33. for”) dx x 35. fx’./5+x2dx .tz—l 2 . dx 2 fo—3x+1 x3—l . d 24 fx‘—-4x x 26. / cosxeSi'” (1x 28. / sec2 item“ dx 30. [cos(2x — Del): 32. / sinJ x cos x dx dx M f(x—3)ln(x-—3) 36. /./1+lnx!n—x-dx X In Problems 37-42, a, b, and c are constants and g(x) is a continuous function whose derivative g’(x) is also continuous. Use substitution to evaluate the indefinite integrals. 2a.: + I) f 1 - ——————~—-—- 38. dx 37 fax2+bx+cdx ax+b 40. fg’ct) sin[g(x)]dx 39. fg’(x)[g(x)]"tlx g’(x) [3(x)12+1 " 41. fg’(x)e"'m dx 42. I 7.1.2 In Problems 43—58, use sttbstitution to evaluate the definite integrals 3 43. f xez +1t1x 2 44. f x5./x3 +2dx l 0 3 2x + 3 2 2x 4 . ————- d 46. —— dx 5 2 (x2 + 3.x?)3 x [a (4x2 + 3W3 s 2 2/2 |n7 ex 47. x — 2 e'(" ’ dx 48. dx ./2‘ ( ) In4 (ex " 3)2 tr/3 .‘r/6 49. / sinx cos x dx 50. / sin2 x cosx dx 0 _ Ir/6 rt/4 :r/J ' 51. f tan x sec2 x dx 52. / smzx dx 0 0 cos x 9 2 53. f x dx 54. f x dx 5 x _ 3 n I + 2 2 t d 2 55. f x 56. —-—{-‘—1—x—————- , .t(ant)2 l (x2 + i) [not2 + 1) 9 1 2 ~’ 57. ——e-fidx 53. f x./4—x2dx 1 fl 0 59. Use the fact that cosx cotx = . srnx to evaluate [cotxdx I 7.2 Integration by Parts and Practicing Integration 1 7.2.1 Integration by Paris As mentioned at the beginning of this chapter, integration by parts is the product rule in integral form. Let u = u(x) and v = v(x) be differentiable functions. Then, differentiating with respect to x yields or, after rearranging, (uv)' = u'v + uv' uu’ = (uv)' -— u’v Integrating both sides with respect to x, we find that fut/(Ix =/(uv)’dx ~ fu'vdx ‘ Since uv is an antiderivative of (uv)’, it follows that f(uv)’cl.t = no + C fr-- 342 Chapter7 l Integration Techniques Combining our results, we find that fx‘/Zln(xme‘)dx =éfx1/Zlnxdx +fx3/2dx 2 = gxslz (lnx -' + BIS/2'4” C onstants. We could denote the integration c + C2, but since Note that we used the same symbol C to mbined them into C = C1 have called them C1 and C2 and then co they stand for arbitrary constants, we need not keep track and can simply capture them all by the same symbol. However, mind that they are not all the same. of how they are related we should keep in Section 7.2 Problems. First, write I 7.2.1 cos2 x = (cos x)(cosx) In Problems 1—30, use integration by parts to evaluate the integrals. 1 f x cosx dx 2 I 3x cosx dx and integrate by parts to show that 2 = - t 2 4. f3xcos(4_x)dx fcos xdx srnxcosx+f51n xdx 3. [Lt cos(3x —- 1) dx Then, use sin2 x + cos2 x = 1 to replace sin2 x in the integral on the right-hand side, and complete the integration of f cos2 x dx. f sin2 x dx sin2 x = (sin x)(sin x) 5. I 2x sin(x — 1) dx 7. fxe" dx 9. fxze" dx 11. fxlnxdx 8 f _ 32. Evaluating the integral requires two steps. First, write 2 2 13' f xlnGxMx 14' f x lnx dx and integrate by parts to show that 2 2 15. fxsec xdx 16. fxcsc xdx [Sinzxdx =_Sinxcosx+fcoszxdx Then, use sin2 x + cos2 x = 1 to replace cos2 x in the integral on : M4 18. I 2x cos x dx 0 the right-hand side, and complete the integration of f sin2 1: dx. 7r/3 17. f xsinxdx o 33. Evaluating the integral 2 e 19. f lnx dx 20. f In x2 dx i 1 4 4 . 21. f in./§ dx 22. f J} 111.5 dx f “‘5‘”de l 1 l 3 requires two steps. 23 xe“ dx 24. f x2e" dx (3) Write 0 w 0 "/6 arcsin x = 1 ~ arcsin x 25 ex Sin x dx 26. f ex cos x dx and integrate by parts once to show that 0 0 27 e-” cos 1x dx 28. -1. . i d f arcsinx dx = x arcsinx — f x dx 2 e Sin 2 x /1 - x2 29. I sing“ x) dx 30. / c050“ x) dx (b) Use substitution to compute 31. Evaluating the integral [ 1x 2 dx (7,7) . __ x ‘ [ cos2 x dx requires two steps. and combine your result in (a) with (7.7) to complete the. computation of f arcsinxdx. :7 7.2 I Integration by Parts and Practicing Integration 34. Evaluating the integral / arccosxdx _ requires two steps. (3) Write arccosx = 1 - arccosx and integrate by parts once to show that x / arccosx dx = x arccosx + f 2 dx - x r (b) Use substitution to compute (7.8) f x d): ‘/l -—x2 and combine your result in (a) with (7.8) to complete the computation of f arccosx dx. 35. (it) Use integration by parts to show that, for x > 0. 1 [llnxdx=(lnx)2—f—lnxdx X X (b) Use your result in (a) to evaluate f—l-lnxdx x 36. (I) Use integration by parts to show that fx"e" dx = x"e" — n ff“? dx Such formulas are called reduction formulas, since they reduce the exponent of x by 1 each time they are applied. (b) Apply the reduction formula in (a) repeatedly to compute fitae‘r dx 37. (I) Use integration by parts to verify the validity of the reduction formula 1 n fx”e‘" dx = -x"e‘" — — x" 1e“ dx a a where a is a constant not equal to 0. (I!) Apply the reduction formula in (a) to compute / xze‘3’r dx 33. (I) Use integration by parts to verify the validity of the reduction formula [(lnx)" dx = .t(ln.\:)’l — n I/‘(lnx)"'l dx (5) Apply the reduction formula in (a) repeatedly to compute / (lnx)3 dx 343 In Problems 39—48, first make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals 39. / cos J; d): 40. 1 sin fidx 41. f x3e“1/2dx 42. / x5e"2dx 43. / sinxcosxe‘i’” dx 1 45. f e‘fi dx 0 . _~ 2 44. [smxcos3xe' "“ ‘dx 2 46. f eV Jr“(1): l 4 1 47. f 1n(./E+1)dx 48. / x3ln(x2+1)dx I 0 I 7.2.2 In Problems 49—60, use either substitution or integration by parts to evaluate each integral. 49. / xe‘z‘ dx 5.. 1.2... tan 1: 53. f 2.): sin(x2) dx 1 55.f16+x2dx x 7. 5 fx+3dx x 59. [XI +3 dx 61. The integral 50. / xe'z‘zdx 52. /———1———dx cscxsecx S4. [22:2 sinx dx 1 56. fo+5dx 1 58. fxz+3dx x+2 60. lxz+2dx flnxdx I can be evaluated in two ways. (I) Write lnx = 1 - lnx and use integration by parts to evaluate the integral. (b) Use the substitution u = lnx and integration by parts to evaluate the integral. 62. Use an appropriate substitution followed by integration by parts to evaluate 2 f x3e“ ’2 dx 63. Use an appropriate substitution to evaluate /x(x - 2)"4 dx 64. Simplify the integrand and then use an appropriate substitu- tion to evaluate /' sin2 x - cos2 x -.——-— dx (smx —- cos x)2 In Problems 65—70, evaluate each definite integral. 4 65. / efi dx 1 /° 2 d x _ l + .r2 4 l N/ e‘sinxdx n 2 66. f ln(x2e‘)dx 1 2 51. 68. f lenxdx I 69 3/6 70.] (1+tan2x)dx 0 350 Chapter7 I Integration Techniques We conclude this section by providing a summary of the two most important cases: when the integrand is a rational function for which the denominator is a e polynomial of degree 2 and is either (1) a product of two not necssarily distinct linear factors or (2) an irreducible quadratic polynomial. The first step is to make sure that the degree of the numerator is less than the degree of the denominator. If not, then we use long division to simplify the integrand. We will now assume that the degree of the numerator is strictly less than the degree of the denominator (i.e., the integrand is a proper rational function). We write the rational function f (x) as P(x) Q(x) with Q(x) = axz + bx + c, a 7e 0, and P(x) := rx + s. Either Q(x) can be factored into two linear factors, or it is irreducible (i.e., does not have real roots). f(x) = Case In: Q(x) is a product of two distinct linear factors. In this case, we write Q(x) = a(x — xi)(x — x2) i i i where x1 and x2 are the two distinct roots of Q(x). We then use the method of partial fractions to simplify the rational function: Q(x) _ ax2+bx+c — a + x—x1 x—xz P(x>_ rx+s 1[ A B ] The constants A and B must now be determined as in Example 3. Case 1b: Q(x) is a product of two identical linear factors. In this case, we write Q(x) = W — x02 where x1 is the root of Q (x). We then use the method of partial fractions to simplify the rational function: Q(x) _ax2+bx+c ~a + x —X1 (x —x1)2 P(x)_ rx+s 1[ A B ] The constants A and B must now be determined as in Example 5. Case 2: Q(x) is an irreducible quadratic polynomial. In this case, Q(x) =ax2+bx +c with b2 ——4ac < 0 and we must complete the square as in Example 2. Doing so then leads to integrals of the form f dx or f x d x2 +1 x2 + 1 x The first integral is tan“ x + C, whereas the second integral can be evaluated by substitution. (See Examples 6 and 7.) Section 7.3 Problems l 7.3.1, 7.3.2 In Problems 5-8, write out the partial-fraction decomposition of the In Problems [—4, use long division to write f (x) as a sum of a fil’lc’ion f (x )- polynomial and a proper rational function. 5 f“) = 2x — 3 6 f”) _ _ x + 1 2x2+5x—1 x2—4x—1 x(x+1) ' — (2x+1)(x-1) 1- f(x)=-—————- 2. f(x)=——————-— A x + 2 x — 1 3x3+Sx—-2x2—-2 x3—3x2—15 4x2—l4x—6 l6x—6 3. (x)=--——— 4. x)=-—————-——— 7. x=—-————-—————-— 8. =————-——— f x2+1 f( x2+x+3 fl ) x(x ~3)(x+1) fa) (2x ~5)(3x+1) 7.4 I lmproperlntegrals 351 . . 2 A 3 In Problems 9—12, write out the partial-fraction decomposrtton of 35' f 1‘2 + 4 dx 36‘ f 7%.? dx the function f (x). x — 4 x _ x + 5x — 1 10 for) _ 9x " 7 In Problems 37-44, evaluate each definite integral. 9. f(x)=x2_1 . ————-—Zx2_7x+3 sx—l 5 x 37. dx 38. / dx ._ “+1 12 f(x)_______12____ 3 x 3X‘1 "'f(x)_x1—3x~10 ' ‘ 3x2+8x-3 1 t 2x2+1d . . . ' 40. x In Problems 13—18, use partial-fraction decomposition to evaluate 39 j; X: + 1 dx /; x the integrals 3 1 3 1 d 1 1 4]. dx 42. f x . d 14. f —— dx 1 _ x 1 _ x2 13 [Ax—2) x x(2x+1) 2| 21 l 1 43. / tan" xdx 44. f x tan" x dx , __ d 16. f —————— dx 15 [emu—3) x .(x—1)(x+2) 0 0 x2 _ 2x _ 2 4x2 .. x _. 1 In Problems 45—52, evaluate each integral. [7. / —-——-— dx 18. f —T—- dx 1 1 x2(X + 2) (x +1) (x _ 3) 45. [(—T (1.: 46. [W dx x x — In Problems 19-22, use partial-fraction decompositon to evaluate x + )4x 2x2 + 2x _ 1 each integral. 47. / __.___ dx 48. f __3____ dx x3-x2+x—4 x3—3x2+x-6d (1“)(1‘l‘xlz x(x“3) . —————————- d 20. ~———————-—————-— x 1 19 f (32 1)(X1 + x (x2 + 2)(X2 + 49. f 2 l 9): dx 50. [m (1X 21/2x2—3x+2d 22 f3x2+4x+3dt 0‘ l x 1 ' 2 x . 2 2 ' . / —-—-—-— d 52. / —-——-—~—-— (1 (x2 +1) (x +1) 51 x2(x2 +1) x (x +1)2(x2 +1) x In Problems 23—26, complete the square in the denominator and 53' (a) To complete Example 8, show that evaluate the integral. 4 4 1 ' 1 x (I ‘ x) o 5 4 2 4 —-—— ——-——————-— ——————=x —-4x +5x ~4x +4— n'/.r2—2x+2dx 24' f‘x1+4x+5dx 1+1:2 1+x2 1 i 1 —— . —— d zs'fx2—4x+13dx 2‘5 fx2+2x+5 x (b) Sh°wmat In Problems 2:46, evaluate each integral. 2x 1 f! 14(12— x)“ dx 5 f' «34:14:- :Y dx 5 [1 x4“ _ x)4 dx _ x o 27. / ———-———dx 28. ——~————————-—— dx 0 0 (x l— 3)“ + 2) (x :- 4)“ + I) and conclude that 29./x2_9dx 30.]x2+9dx 1 <3 n< 1 1 1 1260 " 7 ' 630 31. dX 32. f 7—7 dx x ‘2" _ 2 x3 - x + Use this result to show that x + l x + 1 33' fx—2+3x+2dx 34' fxz+3dx 3.14052: 53.142 Ll 7.4 Improper Integrals In this section, we discuss definite integrals of two types with the following characteristics: M..~.._.‘:- ' 1. One or both limits of integration are infinite; that is, the integration interval is unbounded; or emit-33'.“ 2. The integrand becomes infinite at one or more points of the interval of integration. We call such integrals improper integrals. I 7.4.1 Type 1: Unboundedlntervals 0 . 0 1 2 3 4 5 " Suppose that we wanted to compute the area of the unbounded region below the Figure 7.9 The unbounded region graph of f(x) = e“ and above the x-axis for x 3 0. (See Figure 7.9.) How would . between the graph of y = e" and we proceed? We know how to find the area of a region bounded by the graph of a the x-axis for x 3 0. continuous function [here, f(x) = e‘-‘] and the x-axis between 0 and z, namely, 2 A(z) =/ e"'dx = —e“‘]3 = l —— e‘z o 362 Chapter7 I integration Techniques Solution The function f (x) = 1/,/x + J)? is continuous on [1, 00). The integrand looks rather complicated, but since x + J)? _<_ x + x for x z 1, it follows that Hence, as shown in Example 2. Therefore, is diVergent. 1 > 1 forx>1 Jx+fi‘\/i? " °° 1 1 . Section 7.4 Problems . l 7.4.1, 7.4.2 All the integrals in Problems 1—16 are improper and converge. Explain in each case why the integral is improper; and evaluate each integral. oo oo 1.] 3e"6‘dx 2.] xe"dx o o 4. [Do dx , Jr(lnx)2 _1 1 6. fwol+x2dx 0° 2 8.] xe—”2dx (X) 00 ¢ 10./ x3e" dx oc °° x 9. ——-—— d L. (1 + x2)2 x 9dx M 12. f d" 1 x lnx n/Z 13. f m” dx 0 u 14.] __di__ ,/sinx —2 (x +1)U3 l 2 dx 15. [llnlxldx W In Problems 17—28, determine whether each integral is convergent. If the integral is convergent, compute its value. “‘1 °° 1 17./; Fdx mdx ‘1 4 1 19.1; Fdx de 21 f2 l d 22 f2 1 d ‘ o(x—1)”’ x ' 0 (x—lr‘ x no 23.] 1 dx 0 x+1 W 25. / dx , .r In x 11. o 1 dx +1 0 24. f. .r—x 26.] dx 1 xlnx Zxdx 2 l 3 . ——-——-— 2. d ,27 Luz-1W 8 L..le " 29. Determine whether N 1 [mx2_1dx is convergent. Hint: Use the partial—fraction decomposition 1 _1 1 1 x2-1_2 x~1 x+1 30. Although we cannot compute the antiderivative of f (x) e'xz/z, it is known that 00 f 5‘2” dx = «27: Use this fact to show that 00 f x2e“"2/2 dx = J27! 6° Hint Write the integrand as 2 x - Dre" ’2) and use integration by parts. 31. Determine the constant c so that m . / ce‘” dx = 1 o 32. Determine the constant c so that W C d = [501+x2 x 1 33. In this problem, we investigate the integral °° l f — dx 1 I” for0<p<oo. 7.4 I Improperlntegrals 363 (a) Forz > 1,set 37. (a) Show that l l _ __ 1 A(z)—/i xpdx l Zfl>0 1 + x2 2‘ and show that 1 f = _,,+. _ 1 . orx 2 1. A“) l -— p(z ) (b) Use your result in (a) to show that for p sé 1 and W l A z) = lnz / dx ( t \/1 + .r2 for p = 1. . ‘ (b) Use your results in (a) to show that, for 0 < p 5 1, IS divergent. “m A“) =00 38. (a) Show that 2"00 1 > 1 > 0 (c) Use your results in (a) to show that, for p > 1. - J27; 1 lim A(z) = —— for x _>, 1. z—wo P ‘7 1 (1)) Use your result in (a) to show that 34. In this problem, we investigate the integral °° 1 ——--—dx 11d [m -—x o x” is divergent. for0<p<oo. In Problems 39—42, find a comparison function for each integrand (:1) Compute and determine whether the integral is convergent. 1 00 00 __ 1 [x]; dx 39.] e“'2/2dx 40.] dx _ 1 oo 1 o for 0 < p < oo. (Hint: Treat the case where p = 1 separately.) 0”: 1 Tx (b) Use our result in a) to compute 41. / dx 42. / dx y ( I I I r———l+x _me..+e_. / _ dx 43. (a) Show that p lnx ‘ x lim —— = 0 for0<c< 1. .v—>oo J; (c) Use your result in (b) to show that (b) Use your result in (a) to show that f’ 1 dx _ 1 Zlnx 5 .5 (7.17) 0 x” l _ p for sufficiently large x. Use a graphing calculator to determinejust forO < p < 1. how large x must be for (7.17) to hold. (d) Show that (c) Use your result in (b) to show that ‘ 1 d on 0 3 x / e-fidx (7.18) o is divergent for p 2 1. I converges Use a graphing calculator to sketch the function 1 7.4.3 for) = e‘fi together with its comparison function(s). and use 35. (a) Show that your graph to explain how you showed that the inte ral in 7.18 . g 0 S e—xz S e—x lS convergent. f0” > L 44. (a) Show that (b) Use your result in (a) to show that lim In—x = 0 ram x 00 f [‘2 dx (b) Use your result in (a) to show that, for any c > 0, i is convergent. ex 2 In" 36. (a) Show that for sufficiently large x. 1 1 (c) Use your result in (b) to show that, for any p > 0, 0 < < — . _ ;——l +x4 _. x2 xpe-\ S e—t/Z for x > 0. provided that x is sufficiently large. (b) Use your result in (a) to show that (d) Use your result in (c) to show that. for any p > 0. ‘ [00 l d .30 x He“ (1.: l v 1 + x4 /0 is convergent. is convergent. 370 Chapter? I Integration Techniques y 2 LS 1.6 1.4 L2 1 0.8 0.6 0.4 0.2 0 0 l [.2 1.4 1.6 L8 2 1 Figure 7.39 The trapezoidal rule for édx with n = 5. In Example 3, since f(x) = x2. it follows that f”(x) = 2 and hence K = 2. The error is therefore bounded by 1 12(42) 2 = 0.0104 which is the same as the actual error. In Example 4, since f(x) = l/x, we have |f”(x)| = 2/x3 5 2 for 1 5 x 5 2 (as in Example 2). Hence, with n = 5, the error bound is at most __ 3 2(2 1) = 0.0067 12(52) The actual error was in fact smaller, only 0.00249. As with the midpoint rule, the theoretical error can be quite a bit larger than the actual error. Section 7.5 Problems I 7.5.1, 7.5.2 In Problems 1—4, use the midpoint rule to approximate each integral with the specified value of n. 2 0 1./x2dx,n=4 2.f(x+1)3dx,n=5 1 —1 l rt/Z 3. f e“dx,n=3 4. f sinxdx,n=4 0 0 In Problems 5—8, use the midpoint rule to approximate each integral with the specified value of It. Compare your approximation with the exact value. ‘1 5.] ——dx,rt=4 2 x 4 7. / fidx,rt=4 0 l 6. / (e”—1)dx,n=4 «1 4 2 8.] ~——dx,n=5 2 J)? In Problems 9—12, use the trapezoidal rule to approximate each integral with the specified value of n. 2 9. / dex,rl =4 l l 11.] e" dx,n = 3 0 0 10. / x3dx.n =5 —1 n/Z 12. / sinxdx,n = 4 0 In Problems 13—16, use the trapezoidal rule to approximate each integral with the specified value of n. Compare your approximation with the exact value. 3 l l3./x3dx,n=5 14. / (1—e“)dx,n=4 l —l 2 2 15. f fidx,n=4 16. / ldx,n=5 0 l x 17. How large should n be so that the midpoint rule approxima- tion of 2 f x2 dx 0 In Problems 18—24, use the theoretical error bound to determine how large n should be. [Hint: In each case, find the second derivative of the integrand, graph it, and use a graphing calculator to find an upper bound on I f "(x)|. ] 18. How large should n be so that the midpoint rule approxima« tionof 21 f—dx 1 x is accurate to within 10'3? is accurate to within 10"? 19. How large should It be so that the midpoint rule approxima- 2 f (‘2’: dx' 0 tion of is accurate to within 10“? ' 20. How large should n be so that the midpoint rule approxima- tion of s l f—dt 2 Int is accurate to within 10"? 21. How large should It be so that the trapezoidal rule approximation of l f e" dx 0 is accurate to within 10”? 22. How large should n be so that the trapezoidal rule approximation of 2 / sinx dx 0 is accurate to within 10“? 7.6 I The Taylor Approximation 371' is accurate to within 10"? 25. (a) Show graphically that. for n = 5, the trapezoidal rule overestimates, and the midpoint rule underestimates, l f dex 0 In each case, compute the approximate value of the integral and compare it with the exact value. (b) The result in (a) has to do with the fact that y = x3 is concave up on [0, 1]. To generalize that result to functions with this concavity property, we assume that the function f (x) is continuous, nonnegative, and concave up on the interval [a. b]. Denote by M,I the midpoint rule approximation, and by 7}. the trapezoidal rule approximation. of LI f (x) dx. Explain in words why [2 Mn 5] f(X)dx s Tn (c) If we assume that f (x) is continuous, nonnegative, and concave down on [a, b], then 23. How large should It be so that the trapezoidal rule approximation of 2! e [-dt ‘t is accurate to within 10"? b M. 2/ f(x)dx 2 T. Explain why this is so. Use this result to give an upper and a lower bound on 24. How large should n be so that the trapezoidal rule approximation of 2 COS X / dx 1 x . Ala/de when n = 4 in the approximation. 1 7.6 The Taylor Approximation 'l 0 1 2x Figure 7.40 The graph ofy = e“ and its linear approximation at 0. In many ways, polynomials are the easiest functions to work with. Therefore, in this section we will learn how to approximate functions by polynomials. We will see that the approximation typically improves when we use higher-degree polynomials. 1 7.6.1 Taylor Polynomials In Section 4.8, we discussed how to linearize a function about a given point. This discussion led to the linear, or tangent, approximation. We found the following: The linear approximation of f (x) at x = a is L(x) = fun + f’(a)(x — a) As an example, we look at f (x) = e‘ and approximate this function by its linearization at x = 0. We find that L(x) = f(0) + f’(0)x = 1 +x (7.19) since f ’(x) = e‘ and f (0) = f’(0) = 1. To see how close the approximation is. we graph both f (.r) and L(x) in the same coordinate system. (The result is shown in Figure 7.40.) The approximation is quite good as long as x is close to 0. The figure suggests that it gets gradually worse as we move away from O. In the approximation, we required only that f (x) and L(x) have in common f (0) = L(0) and f’ (0) = L’(0). MQ-‘nnn‘LW—rm‘Am—J w».=.‘., A A - - - -.=—........« - - A .s . A: a low. a— eflfi—a—ec; “ : I” I. . ., ~ set—A; ; M .-=\ n in.“ WWW—#6“ AA...~....‘. . g .g. <~_. ..._. . A. v. 7.6 l The Taylor Approximation 381 a polynomial of degree 7 might not. We can easily check this; we find that 1 1 1 1 1+1+Z+§+Z+...+7i=2.71825396825 1 1 1 1 l 1“271827876984 + +§i+§i+fi+n-+8—!— . Comparing these with e = 2.71828182846 . . . , we see that the error is equal to 2.79 x 10’5 when n = 7 and 3.06 x 10'6 when n = 8. The error that we computed with (7.28) is a worst—case scenario; that is, the true error can be (and typically is) smaller than the error bound. I . We have already seen one example in which a Taylor polynomial was useful only for values close to the point at which we approximated the function, regardless of n, the degree of the polynomial. In some situations, the error in the approximation cannot be made small for any value close to the point of approximation, regardless of n. One such example is the continuous function e‘l/x forx > 0 0 forx 5 0 - l which is used, for instance, to describe the height of a tree as a function of age. We can show that f (W0) == 0 for all k z 1, which implies that a Taylor polynomial of degree n about x = 0 is 3.06) = 0 for all n. This example clearly shows that it will not help to increase n; the approximation just will not improve. When we use Taylor polynomials to approximate functions, it is important to know for which values of x the approximation can be made arbitrarily close by choosing n large. Following are a few of the most important functions, together with their Taylor mim'xihh, 3; ~ A " 'wu A c U .r-r 4: - z r “4.; A ~=. - “u - ‘r "w w A A 1“ A polynomials about x = 0 and the range of x values over which the approximation can be made arbitrarily close by choosing n large enough: 2 3 n ‘ ‘ I x x x e =1+x+—+-+---+—-+Rn+1(x),~oo<x<oo 2! 3! n! _ x3 x5 x7 x9 x2n+1 smx=x-—§+§—-i+fi—~a+(—l) m+Rn+1(X), —oo<x <00 x2 x4 x5 1c8 x2" :1... .__-_ _.._... -1"__ n ,_ cosx 2!+4! 6!-l~8! +( )(Zn)!+R+1(x) oo<x<oo x2 x3 x4 x5 x" ln(1+x) =x— E-+?—7+?-—---+(—1)"+1;-+R,,+1(x), -1 <x SI 1 1_x =1+x+x2+x3+x4+---+x"+R,.+1(x), ~1<x <1 Section 7.6 Problems . I 7.6.1 In Problems 6-10, compute the Taylor polynomial of degree n 'n Problems 1—5, find the linear approximation of f (x) at x = 0. about a = of” the indicatedfi‘m’iom' l- ftx) = ez‘ 2- f(x) = sin(3x) 6. f(x) = I: ,n = 4 7. f(x) = cosx,n =5 1 _ ‘ x "f(‘)=1_; 4400-" 8.f(x)=e3‘,n=3 9. f(x)=.r5,n=6 i. f(x)=ln(2+x2) 10. f(x)=‘/1+x,n=3 382 Chapter7 1 Integration Techniques ln Problems 11—16, compute the Taylor polynomial of degree n about a = 0 for the indicated functions and compare the value of the functions at the indicated point with the value of the corresponding Taylor polynomial. ll. f(x) = ,/2+x,n = 3,x =0.l l 12- f (x) = l _ x 13. f(x) §sinx,n =S,x =1 14. f(x) = e“,n = 4,1: = 0.3 15. f(x) = tanx,n = 2,1: =0.1 16. f(x)=ln(1+x), n = 3,x = 0.1 17. (a) Find the Taylor polynomial of degree 3 about a = 0 for f(x) = sin x. - (b) Use your result in (a) to give an intuitive explanation why ,n=3,x=0.l ‘ sinx hm ——- = 1 .r—>0 1' 18. (:1) Find the Taylor polynomial of degree 2 about a = 0 for f (x) = cos x. (b) Use your result in (a) to give an intuitive explanation why cosx —1 lim _ 0 x—>0 x l 7.6.2 In Problems 19—23, compute the Taylor polynomial of degree n about a and compare the value of the approximation with the value of the function at the given point x. 19. f(x) = Jim = l.n =3;x =2 20. f(x) = lnx.a = Ln = 3;x = 2 21. f(x)=cosx,a= ig-,n=3;.\t=-’7L 22. f(x) = x'I’,a = -1,n = 3;x = —o.9 23. f(x) = e’,a =2,n =3;x =2.1 24. Show that r“ -. r; + 43% — T,) for T close to T... 25. Show that, for positive constants r and k. N rN(1—-k-—)~rN for N close to 0. 26. (a) Show that, for positive constants a and k aR a ~" f(R)=k+R i R for R close to 0. l 7.7 Tables of Integrals (Optional) (b) Show that, for positive constants a and k. aR a a = 9: — — R — k f(R) k+R 2+4k( ) for R close to k. I 7.6.3 In Problems 27—30, use the following form of the error term _ f WW) n+1 Rn+l(x) - (n +1”): where c is between 0 and x, to determine in advance the degree of Taylor polynomial at a = 0 that would achieve the indicated accuracy in the interval [0, x]. (Do not compute the Taylor polynomial.) ' 27. f(x) = e‘, x = 2, error < 10" 28. f(x) = cosx. x = 1, error < 10‘2 29. f(x) = 1/(1+ x). x = 0.2, error < 10‘2 30. f(x) = ln(1+x),x = 0.1, error < 10‘2 31. Let f(x) = e“" forx > Oand f(x) = 0 forx = 0. Compute a Taylor polynomial of degree 2 at x = 0, and determine how large the error is 32. We can show that the Taylor polynomial for f (x) = (1 + x)“ aboutx = 0, with a a positive constant, converges for x e (—1, 1). Show that ot(at—1)2 2! x + a(at ‘13)'(a — 2)x3 + . . . + Rn+l(x) (1+x)"=l+ax+ 33. We can show that the Taylor polynomial for f (x) = tan‘l x about at = 0 converges for Ix] _<_ 1. (a) Show that the following is true: x3 5 7 -1 X x tan X=X—-3-+§-—.7—-+---+Rn+l(x) (b) Explain why the following holds: a l K — 1 - 3- + +... \Hu— 1 S (This series converges very slowly, as you would see if you used it to approximate It.) Before the advent of software that could integrate functions, tables of indefinite integrals were useful aids for evaluating integrals In using a table of integrals, it is still necessary to bring the integrand of interest into a form that is listed in the table— and there are many integrals that simply cannot be evaluated exactly and must be ‘ evaluated numerically. We will give a very brief list of indefinite integrals and explain how to use such tables Chapter 7 l Review Problems 387 I 7.7.1 A Note on Software Packages That Can Integrate Mathematicians and scientists use software packages to integrate functions. They are ' not difficult to use with some practice. Although they will not give you insight into what technique could be used to solve an integration problem, they quickly give you the correct answer. For instance, if we used MATLAB, one of the common software packages, to calculate the integral in Example 8 of Section 7.3, we would enter the following string of commands into our computer: syms x; f-x‘ 4* (l-x) ‘ 4/(1+x‘ 2); int(£,0, 1) MATLAB then returns ans - 22/7-pi Section 7.7 Problems In Problems 1—8, use the table on pages 383—384 to compute each 11. f (x2 _ Ue—x/Z dx n f (x + Dze—Zx dx integral. x dx 2 . 2. x l / 2x __ 3 dx / 16 + x2 13. /c0s2(5x - 3) dx 14. fm dx I. f «x2 — 16dx 4. / sin(2x) cos(2x) dx 1 n“ 15. fv9+4x2dx 16. S. / x3e" dx 6. f e“ cos(2x) dx _ x o o . .2 dx 17. f e2“1 sin (5x) dx 18. f (x —1)2e2"dx 7. f x2 lnx dx 8. 2 . . 1 , x lnx 4 1 t ’n Problems 9—22, use the table on pages 383—384 to compute each 19. / dx 20. f (x + 2)2 111x dx ntegral after manipulating the integrand in a suitable way. 2 x In J; 1 n/6 ” 2 ' 3 l. A e cos (x — dx 10. [I x ln(2x — 1) dx 21. fcos(ln(3x)) dx 22. f m dx Chapter 7 Key Terms )iscuss the following definitions and 7. Proper rational function 14. Numerical integration: midpoint and onceptx 8. Irreducible quadratic factor trapezoidal rule . The substitution rule for indefinite 9. ImprOper integral 15. Emr bounds for the midpoint and “egrfils 10. Integration when the interval is the trapezoidal rule . The substitution rule for definite unbounded 16 U _ b , l ltegrals ll. Integration when the integrand is _ t‘ 5:93 ta 1“ 0f “mg” 3 for - - in e ra ion . Integration by parts Esc‘g‘tmuous d d. f g ‘ ' I . The “trick” of “multiplying by 1” . ' onvérgence an wergence 0 17' Lmear approx’manon improper integrals . . Partial-fraction decomposition 13. Comparison results for improper 18' Taylor POlynomlal 015 degree n . Partial-fraction method integrals 19. Taylor’s formula ‘ hapter 7 Review Problems lProbIems 1—30, evaluate the given indefinite integrals. 5. (1 + fl) 1/3 dx 6. f x /3x + 1 4,; [x20 — x3)2dx 2. / mi dx 1 + 5‘“ x 7. [x sec2(3x2) dx 8. ftanx seczx dx . 2 [Me—12d): 4. f‘ln(1+x)dx 9. fxlnxdx 10. fx’lnxzdx 1+x2 388 Chapter7 l Integration Techniques 11. [seczx ln(tanx) dx 1 13. f4+x2dxl 15. / tanxdx 17. [en sinxax (x + l)2 d x-l x 12. ffilnfidx l 14. f4_x2 dx 16. ftan"xdx 18. [x sinx dx 20. finfidx 22. [sinx cosx e’i'” dx 1 . ——————d 2" forum—2) " x 26. [X2 + 5 dx 1 2&fx21+5dx 30.] 2x+1 dx .ll—xI In Problems 31—50, evaluate the given definite integrals. 321 31.fx+dx 1x I 33. f "4/2 dx 0 l 47. f xlnxdx 0 tr/4 49. / em” sinx dx () tr/2 32. / xsinxdx u 2 34. / lnxdx 1 1/2 2 36. / dx 0 ‘/1—-J\:2 °° 1 40./0. x2+3dx °° 1 4 1/0 (xi-3)”: °° 1 44.] 7d}! 1 X on 1 46.] ——--dx 1 J? l 48. f 152‘ dx 0 3/4 50. f xsin(2x) dx 0 In' Problems 51—54, use (a) the midpoint rule and (b) the trapezoidal rule to approximate each integral with the specified value of n. 2 I 51.f(x2—l)dx,n=4 52. / (x3—1)dx.n=4 0 -1 l rr/4 53. f e" (Ix, n = 5 54. / sin(4x) dx, n = 4 o 0 In Problems 55-58, find the Taylor polynomial of degree n about x = a for each function. 55. f(x) = sin(2x). a = 0, n = 3 56. f(x) = e"1/2,a = o, n = 3 57. f(x)=lnx,a=1,n=3 58. f(x) = x _ 3, a — 4, n _. 4 59. Cost of Gene Substitution (Adapted from Roughgarden, I996) Suppose that an advantageous mutation arises in a population. Initially, the gene carrying this mutation is at a low frequency. As the gene spreads through the population, the average fitness of the population increases. We denote by fmo) the average fitness of the population at time t, by f.v3(0) the average fitness of the population at time 0 (when the mutation arose), and by K the final value of the average fitness alter the mutation has spread through the population. Haldane (1957) suggested measuring the cost of evolution (now known as the cost of gene substitution) by the cumulative difference between the current and the final fitness—that is, by A (K _ favg(t))dt In Figure 7.43. shade the region whose area is equal to the cost of gene substitution. 1....(0 11.40) t Figure 7.43 The cost of gene substitution. See Problem 59. ...
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