Chapter 3. Techniques of integration. Contents. 3.1 Recap: Integration in one variable. This material is in Chapter 7 of Anton Calculus.

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1 Chpter 3. Techniques of integrtion This mteril is in Chpter 7 of Anton Clculus. Contents 3. Recp: Integrtion in one vrible Antierivtives we know Substitution (reminer) Integrtion by prts Trigonometric Integrls Inverse trigonometric substitutions Prtil Frctions Improper integrls Using computers to fin integrls Trpezoil rule formul Simpson s rule formul Recp: Integrtion in one vrible There re two topics with similr nmes: Reverse of ifferentition Inefinite integrl f(x) x = most generl ntierivtive for f(x) The term ntierivtive for f(x) mens function whose erivtive turns out to be f(x). Definite integrl This is relte to summtion (it is limit of sums of certin kin). The integrl sign ws originlly invente s moifie S (for sum). There is no reson to expect connection between these two ifferent things, but there is. 3.. Theorem (Funmentl Theorem of Integrl Clculus). Assume tht y = f(x) is continuous for x b. Consier A(x) = x f(t) t for x b. (A(x) is new function, built from f n efinite integrtion.) Then A(x) is n ntierivtive for f (tht is A (x) = f(x) for x b). In summry: ( x ) f(t) t = f(x) ( x b, if f continuous) x

2 05 6 Mthemtics MAE0 This is one prt of the Funmentl theorem, or one wy to stte it. 3.. Corollry. There is n ntierivtive for every continuous function f Exmple. Fin ( x ) x t t Solution: By the theorem ( x ) x t t = x Well, wht bout the hypothesis tht the integrn, f(t) = /t in this exmple shoul be continuous? For tht we nee to voi t = 0. We nee to suppose tht 0 is not in the close intervl from to x (or from x to if x < ). We lso nee to ssume tht for x /t t to mke sense, n so perhps you cn sy tht the bove solution must be right becuse the question of fining the erivtive oes not mke sense if 0 is between n x Theorem (Other prt of funmentl theorem). Assume tht y = f(x) is continuous for x b n suppose g(x) is n ntierivtive for f(x) (tht is g (x) = f(x) for x b). Then b This is fmilir. For instnce: 3..5 Exmples. (i) Fin f(x) x = g(b) g()= [g(x)] b x= e x x Solution: For f(x) = e x, we nee g(x) with g (x) = f(x) (n ntierivtive). We cn guess n nswer if we notice tht n so Tht is g(x) = (/)e x works. So 4 x ex = e x x (x) = ex () = e x e x x = x ex = e x [ ] ex = e4 e (ii) Fin x x Solution: We lerne recently tht ln x hs erivtive /x for x > 0. So for f(x) = /x (with x > 0) we cn tke g(x) = ln x. We get 4 x x = [ln x]4 = ln 4 ln

3 Techniques of integrtion 3 We cn simplify tht becuse ln = 0. So 4 x = ln 4 x Asie: Tht woul hve worke with 4 replce by ny positive vlue n in fct b x = ln b if b > 0 x Switching roun some letters, tht mens x t = ln x if x > 0 t An lterntive pproch to where the functions e x n its inverse ln x come from is to strt by efining ln x = x t, use the form Theorem 3.. of the Funmentl Theorem to t show (/x) ln x = /x for x > 0, n then erive the other properties of ln. Finlly get to the exponentil s the inverse. This is one in Anton 6.6. Notice tht to use Theorem 3..4 to evlute n integrl, you nee ifferent wy to fin n ntierivtive g(x) for the integrn f(x) thn Theorem 3.. or Corollry 3... If you use Theorem 3.. to fin g(x) = A(x), n then use Theorem 3..4, you en up sying tht the integrl is equl to itself. 3. Antierivtives we know (Anton 7..) At the en result of our stuies will be tht we will be ble to fin certin number of integrls using the techniques we will escribe. But, unlike ifferentition where we cn ifferentite lmost nything we cn write own using the bsic rules (incluing the chin rule, prouct rule n quotient rule), with integrtion it is esy to come cross simple-looking things we will not be ble to o. One exmple is sin(x ) x. For ech ifferentition formul, we hve corresponing integrtion formul. Here re ones we know (mostly).

4 Mthemtics MAE0 Derivtive formul x xn = nx n x ln x = x x ex = e x x x = x (x > 0) sin x = cos x x cos x = sin x x x tn x = sec x sec x = sec x tn x x Integrtion formul x n x = n + xn+ + C if n x = ln x + C x e x x = e x + C x x = + x+ + C if, x > 0 cos x x = sin x + C sin x x = cos x + C sec x x = tn x + C sec x tn x x = sec x + C

5 Techniques of integrtion 5 Derivtive formul x sin x = ( < x < ) x x tn x = + x sinh x = cosh x x cosh x = sinh x x x tnh x = sech x x sinh x = + x x cosh x = x tnh x = x (x > ) x Integrtion formul x = x sin x + C + x x = tn x + C cosh x x = sinh x + C sinh x x = cosh x + C sech x x = tnh x + C x = + x sinh x + C = ln(x + x + ) + C x x = cosh x + C = ln(x + x ) + C x x = tn x + C = ln + x + C ( < x < ) x x x = ln + x x + C The lst of these formule is not quite one we knew lrey. It works for x <, for < x < n for x >. Also the one tht sys (/x) x = ln x + C nees some explntion. It is true for x > 0 (we know (/x) ln x = /x if x > 0 n x = x if x > 0. For x < 0, we hve x = x n then ln x = x x ln( x) = ( ) = x (x < 0). x So the formul is correct for either cse x > 0 or x < 0 but in fct it is not permissible to cross from x > 0 to x < 0 in the sme iscussion (becuse /x oes not mke sense t x = 0 n efinite integrls of /x lso won t mke sense if x = 0 is inclue in the integrtion). Aprt from these bsic integrls, there re integrtion formule tht follow from the chin rule for ifferentition n the prouct rule for ifferentition. Essentilly, when we rerrnge the integrl of the two formule, we get the methos of substitution n integrtion by prts.

6 Mthemtics MAE0 3.3 Substitution (reminer) (Anton 4.3.) Lst semester you stuie the technique of substitution, which is in essence bse on the chin rule. We recll some simple exmples: 3.3. Exmples. (i) Consier e x x (which rose erlier n we solve more or less by guessing ). A simple substitution tht works here is u = x, leing to u = x, or x = u n e x x = e u u = eu u = eu + C = ex + C (ii) Consier the slightly more complicte problem of fining x cos(x 3 ) x Solution: If we notice tht u = x 3 hs u x = 3x n is fctor in the integrn prt from the constnt fctor 3, then we cn write x cos(x 3 ) x Let u = x 3 x cos(x 3 ) x = = u x = 3x u = 3x x u 3x = x 3x cos u u 3x cos u u 3 = 3 sin u + C = 3 sin(x3 ) + C For substitution to work, it is usully necessry for u/x to be fctor in the integrn (essentilly, mybe in slightly hien wy). It is vitl tht we express ll x s n x s in terms of u n u.

7 Techniques of integrtion Exmple. For efinite integrl like 5 x cos(x 3 ) x, where substitution is useful, we cn work out the inefinite integrl s bove n then put in the limits, or we cn chnge to limits for u like this 5 x cos(x 3 ) x = u=5 3 u=5 [ ] u=5 cos u u = cos u u = u= 3 3 u=8 3 3 sin u = u=8 3 sin 5 3 sin 8 To reiterte, if we mke substitution u = u(x) in efinite integrl b f(x) x we cn chnge it to new efinite integrl where the limits re the limits for u tht correspon to x = n x = b: b x=b u=u(b) f(u(x)) x = f(u(x)) x = f(u) x u u x= u=u() (n this hs the vntge tht is sys one number is equl to nother no nee to go bck n express u in terms of x when you get the ntierivtive, something tht cn be messy in some exmples). 3.4 Integrtion by prts (Anton 7..) If we integrte both sies of the prouct rule we get or u (uv) = x x v + uv x (uv) x = x uv = v u x x + v u x x + u v x x u v x x This llows us wy of trnsforming integrls tht tke the form of prouct of one function times the erivtive of nother u v x = uv v u x x x into ifferent integrl (where the ifferentition hs flippe from one fctor to the other). The vntge of this comes if we know how to mnge the new integrl (or t lest if it is simpler thn the originl). The integrtion by prts formul is usully written with the x s cncelle u v = uv v u

8 Mthemtics MAE Exmples. () x ln x x Solution: The two most obvious wys to use integrtion by prts re u = x, v = ln x x (Problem with this is we cn t fin v very esily) u = ln x, v = x x It turns out tht the secon is goo. x ln x x Let u = ln x v = x x u = x x v = x x ln x x = u v = uv v u = (ln x) x x x x x = (x /) ln x x = (x /) ln x x 4 + C (b) e ln x x Solution: This is one of very few cses which cn be one by tking v = x n u = the integrn. Every integrl tkes the form u v in tht wy, but it is rrely goo wy to strt integrtion by prts. Here we cn mke use of the efinite integrl form of the integrtion by prts formul b u v = [uv] b which rises in the sme wy s the inefinite integrl formul (tke efinite integrls of the prouct rule for ifferentition). b v u

9 Techniques of integrtion 9 e e ln x x Let u = ln x v = x u = x x v = x ln x x = e u v = [uv] e e = [(ln x)x] e v u e = e ln e ln e = e [x] e = e (e ) = x x x x (c) x cos x x Solution: x cos x x Let u = x v = cos x x u = x x v = sin x x cos x x = u v = uv v u = x sin x sin x(x) x = x sin x x sin x x The point here is tht we hve succeee in simplifying the problem. We strte with x times trigonometric function (cos x) n we hve now got to x times trigonometric function (sin x this time, but tht is not so ifferent in ifficulty to cos x). If we continue in the sme (or similr) wy n pply integrtion by prts gin, we cn mke the problem even

10 Mthemtics MAE0 simpler. We use U n V this time in cse we might get confuse with the erlier u n v. x sin x x Let U = x V = sin x x U = x V = cos x x sin x x = U V = UV V U = x( cos x) ( cos x) x = x cos x + cos x x = x cos x + sin x + C Combining with the first stge of the clcultion x cos x x = x sin x + x cos x sin x C n, in fct C is plus nother constnt. Since C cn be ny constnt, the nswer x cos x x = x sin x + x cos x sin x + C is lso goo Remrk. We will not in fct lern ny other techniques thn these which re purely integrtion methos. We will spen some time explining how to mke use of these techniques in specific circumstnces (s it is often not t ll obvious how to o so). There is one other metho we will come to clle prtil frctions, metho for integrting frctions such s x + (x )(x + x + ) x However, the thing we hve to lern bout is lgebr wy to rewrite frctions like this s sums of simpler ones n there is no new ie tht is irectly integrtion. The lgebr llows us to tckle problems of this sort. 3.5 Trigonometric Integrls (Anton 7.3.) (i) Powers of cos x times powers of sin x with one power o Metho: For sin n x cos m x x One thing to voi is U = v n V = u becuse this will just unrvel wht we i to begin with.

11 Techniques of integrtion if n = the power of sin x is o, substitute u = cos x if m = the power of cos x is o, substitute u = sin x 3.5. Exmple. sin 3 x cos 4 x x Solution: Let u = cos x, u = sin x x, x = sin 3 x cos 4 x x = = = = = u sin x sin 3 xu 4 u sin x sin xu 4 u ( cos x)u 4 u ( u )u 4 u u 6 u 4 u = 7 u7 5 u5 + C = 7 cos7 x 5 cos5 x + C (ii) Powers of cos x times powers of sin x with both powers even Metho: use the trigonometric ientities sin x = ( cos x), cos x = ( + cos x) 3.5. Exmple. sin 4 x cos x x Solution: sin 4 x cos x x = (sin x) cos x x ( ) ( ) = ( cos x) ( + cos x) x = ( cos x + cos x)( + cos x) x 8 = cos x cos x + cos 3 x x 8

12 05 6 Mthemtics MAE0 Now x is no bother. cos x x is not much hrer thn cos x x = sin x + C. If we look t sin x = (cos x) x we cn see tht cos x x = sin x+c. (This cn lso be one by substitution u = x but tht is hrly neee.) Next cos x x = (using the sme ies s for cos x x). ( + cos 4x) x = (x + 4 ) sin 4x + C For cos 3 x x we re in sitution where we hve n o power of cos times zeroth power of sin. So we cn use the erlier metho (the fct tht the ngle is x oe snot mke big ifference) of substituting u = sin x. Then u = cos x x, x = u cos x, cos 3 x x = cos 3 u x cos x = cos x u = sin x x = u u = (u 3 ) u3 + C = sin x 6 sin3 x + C Putting ll the bits together sin 4 x cos x x = cos x cos x + cos 3 x x 8 = ( x 8 sin x x 8 sin 4x + sin x ) 6 sin3 x (iii) Powers of sin x n cos x = 6 x 64 sin 4x 48 sin3 x + C Metho: Use the previous two methos treting sin n x x = sin n x(cos x) 0 x n similrly for cos m x x (tht is tret the secon power s the zeroth power). + C

13 Techniques of integrtion Exmples. cos 3 x x Solution: cos 3 x x = (sin x) 0 cos 3 x x. Power of cos o. Substitute u = sin x, u = cos x x, x = u cos x cos 3 x x = = = cos 3 x cos x u u cos x u u = u 3 u3 + C = sin x = 3 sin3 x + C sin 4 x x Solution: Use even powers metho. sin 4 x x = (sin x) x ( ) = ( cos x) x = cos x + cos x x 4 Note: still hve one even power = cos x + ( + cos 4x) x 4 = 3 4 cos x + cos 4x x = ( 3 4 x sin x + ) sin 4x + C 8 = 3 8 x 4 sin x + sin 4x + C Inverse trigonometric substitutions (Anton 7.4.) We now consier clss of substitutions which seem quite counter intuitive.

14 Mthemtics MAE0 Recll these x sin x = x tn x = x cosh x = x sinh x = The corresponing substitution methos re: x + x x x + integrls involving x, substitute θ = sin x (or it is often more convenient to write it x = sin θ). More generlly, integrls involving u (with constnt) try substituting u = sin θ (or θ = sin (u/)). integrls involving try substituting u = tn θ + u integrls involving u try substituting u = cosh t integrls involving u + try substituting u = sinh t 3.6. Exmples. (i) 9 4(x + ) x Solution: In this cse, we hve u with = 9, = 3, u = (x + ) n so our metho sys to try u = sin θ or (x + ) = 3 sin θ x = 3 cos θ θ x = 3 cos θ θ 9 4(x + ) x = 9 9 sin θ 3 cos θ θ 9 = cos θ 3 cos θ θ = 3 cos θ 3 cos θ θ = 9 cos θ θ = 9 + cos θ θ 4 = 9 4 (θ + sin θ) + C

15 Techniques of integrtion 5 To get the nswer in terms of x, we nee θ in terms of x (x + ) = 3 sin θ (x + ) = sin θ 3 ( ) θ = sin 3 (x + ) n we coul get correct nswer by replcing θ by this everywhere in the nswer bove. There is wy to simplify the nswer but we won t go into tht. (ii) 4x 6x 7 x Solution: For qurtics insie squre root like this, wht we shoul o first is complete the squre. Tht is, rerrnge the x n x terms so tht (with suitble constnt) we get multiple of perfect squre 4x 6x 7 = 4(x + 4x) 7 = 4(x + 4x + 4 4) 7 = 4(x + 4x + 4) + 9 = 9 4(x + ) This mens tht not only is this problem similr to the previous one, it is in fct the sme problem gin (now tht we complete the squre) Remrk. There re in fct mny more tricks we coul go into. 3.7 Prtil Frctions (Anton 7.5.) Prtil frctions re technique from lgebr, but our reson for eling with them is tht they cn in principle help fin every integrl of the form p(x) q(x) x where p(x) n q(x) re polynomils. Except in few specil cses, we on t yet know how to fin such integrls. One specil cse, where we on t nee prtil frctions, is where q (x) = p(x) or q (x) = αp(x) for some constnt α, becuse in these cses we cn mke substitution u = q(x), u = q (x) x n it will work out nicely. In fct substitution woul lso work if q(x) = r(x) n for some n n r (x) = αp(x) for constnt α we cn substitute u = r(x), u = r (x) x = αp(x) x, p(x) q(x) x = p(x) r(x) n x = p(x) u n u αp(x) = (/α) u u n

16 Mthemtics MAE0 The ie of prtil frctions is to rewrite p(x) s sum of frctions with simple enomintors q(x) n numertors tht re somehow smll compre to the enomintor. We nee to explin exctly how it goes. We nee to tlk bout fctoring polynomils s much s possible. To strt with, polynomil is n expression you get by tking finite number of powers of x with constnt coefficients in front n ing them up. For exmple or p(x) = 4x x + 7 p(x) = 7x + 5x 0 x 9 + x 8 + x + 5 re polynomils. The highest power of x tht hs nonzero coefficient in front is clle the egree of the polynomil. The exmples bove hve egree n egree. Wht is hny to know is tht when we multiply polynomils, the egrees. So (x + )(x + 5)(x + x + ) will hve egree + + = 4 when it is multiplie out. Constnt polynomils hve egree 0, except the zero polynomil we re best not giving ny egree to the zero polynomil. Now, wht kin of polynomil cn be fctore? For this purpose we on t consier constnt fctors s genuine fctors. So x + 4 = (x + ) = 3 (6x + ) will not be counte s fctoriztion. Anything of egree certinly cnnot be fctore then. Some things of egree cn be fctore, such s x + 5x + 4 = (x + )(x + 4) but other qurtics cnnot be fctore if we on t llow complex numbers to be use. We cnnot fctor x + x + = (x α)(x β) becuse if we coul then the roots of x + x + woul be α n β. The roots of x + x + re b ± b 4c = ± 4 8 = ± n these re complex numbers. So α n β woul hve to be these complex numbers. A remrkble fct is tht every polynomil of egree 3 or more cn be fctore, t lest in theory. It oes not men it is esy to fin the fctors, unfortuntely. Wht you cn sometimes rely on to fctor polynomils is the Reminer Theorem. Recll tht it sys tht if p(x) is polynomil n you know root x = (tht is vlue so tht p() = 0), then x must ivie p(x). Using the theory, just s in principle whole numbers cn be fctore s prouct of prime numbers, so polynomils p(x) with rel coefficients cn be fctore s prouct of liner fctors like x n qurtic fctors x + bx + c with complex roots. If the coefficient of the

17 Techniques of integrtion 7 highest power of x in p(x) is not, then we lso nee to inclue tht coefficient. So complete fctoriztion of x + 8x + = (x + 4)(x + ) = (x + )(x + ) = (x + ). For 3x 3 + 3x + 6x + 6 = 3(x 3 + x + x + ), you cn check tht x = is root n so x ( ) = x + must ivie it. We get 3x 3 + 3x + 6x + 6 = 3(x + )(x + ) from long ivision. Now we cn outline how prtil frctions works for frction p(x) of two polynomils: q(x) Step 0: (preprtory step). If the egree of the numertor p(x) is not strictly smller thn the egree of the enomintor q(x), use long ivision to ivie q(x) into p(x) n obtin quotient Q(x) n reminer R(x). Then n egree(r(x)) < egree(q(x)). p(x) q(x) = Q(x) + R(x) q(x) We concentrte then on the proper frction prt R(x)/q(x). Step : Now fctor q(x) completely into prouct of liner fctors x n qurtic fctors x + bx + c with complex roots. Gther up ny repete terms, so tht if (sy) q(x) = (x )(x + )(x + 3)(x + ) we woul write it s q(x) = (x )(x + ) (x + 3). Step : Then the proper frction R(x) q(x) types: (i) (ii) A (x ) m Bx + C (x + bx + c) k cn be written s sum of frctions of the following where we inclue ll possible powers (x ) m n (x + bx + c) k tht ivie q(x). The A, B, C stn for constnts. As exmples, consier the following. We just write own wht the prtil frctions look like. In ech cse, we strt with proper frction where the enomintor is completely fctore lrey. So some of the hr work is lrey one. (i) x + x + 5 (x )(x )(x 3) = A x + A x + A 3 x 3

18 Mthemtics MAE0 (ii) (iii) x 3 + x + 7 (x + ) (x 4) = A x + + A (x + ) + A 3 x 4 x + x + (x + )(x + x + ) = A x + + Bx + C x + x + (iv) x 4 + x + (x + )(x + x + ) = A x + + B x + C x + x + + B x + C (x + x + ) To mke use of these, we hve to be ble to fin the numbers A, B, C,... tht mke the eqution true. Tke the first exmple x + x + 5 (x )(x )(x 3) = A x + A x + A 3 x 3. To fin the pproprite A, A, A 3, we multiply cross by the originl enomintor (x )(x )(x 3). This hs the effect of clering ll the frctions. x + x + 5 = A x (x )(x )(x 3) + A (x )(x )(x 3) x + A 3 (x )(x )(x 3) x 3 = A (x )(x 3) + A (x )(x 3) + A 3 (x )(x ) There re two venues to pursue from here. In this cse, metho seems esier to me, but in generl metho cn be s goo. Metho. Plug in the vlues of x tht mke the originl enomintor (x )(x )(x 3) = 0. x = : = A ( )( 3) + A (0) + A 3 (0) 7 = A A = 7/ x = : = A (0) + A ()( ) + A 3 (0) = A A = x = 3 : 7 = A 3 ()() A 3 = 7/ So we get x + x + 5 (x )(x )(x 3) = 7/ x + x + 7/ x 3.

19 Techniques of integrtion 9 Our interest in this is for integrtion. We cn now esily integrte x + x + 5 7/ (x )(x )(x 3) x = x + x + 7/ x 3 x = 7 Metho. Multiply out the right hn sie. ln x ln x + 7 ln x 3 + C x + x + 5 = A (x )(x 3) + A (x )(x 3) + A 3 (x )(x ) = A (x 5x + 6) + A (x 4x + 3) + A 3 (x x + ) = (A + A + A 3 )x + ( 5A 4A A 3 )x + (6A + 3A + A ) n compre the coefficients on both sies to get system of liner equtions A + A + A 3 = 5A 4A A 3 = 6A + 3A + A = 5 These cn be solve (by Gussin elimintion, for exmple) to fin A, A, A 3. Metho is certinly mgic in this cse, but there re exmples where Metho oes not get ll the unknown so esily. Another exmple. Here is one of our previous exmples with the numbers worke out. x + x + (x + )(x + x + ) = 0x + x + x + x + To fin the integrl of this, x + x + (x + )(x + x + ) x = x + x = ln x + 0x + x + x + x 0x + x + x + x To work out the remining integrl, we use the metho of completing the squre x + x + = x + x + + = (x + ) + n there is trick. The trick is inspire by the fct tht the substitution u = x + x +, u = (x + ) x = (x + ) x woul work if the numertor ws multiple of x +. Wht we cn o is write 0x + x + x + x = 0x + 0 x + x + x + x + x + x n mke the u = x + x + substitution in the first hlf, while the secon is n inverse tn exmple. (By substituting w = x +, w = x, the secon prt becomes (x + ) + x = w + w = tn w = tn (x + )

20 Mthemtics MAE0 or we might be ble to guess tht.) We get 0x + 0x + 0 x + x + x = u Finlly, our integrl works out s u (x + ) + 5 = u u + tn (x + ) = 5 ln u + tn (x + ) + C (x + ) + x = 5 ln(x + x + ) + tn (x + ) + C x + x + (x + )(x + x + ) x = ln x + 5 ln(x + x + ) tn (x + ) + C 3.8 Improper integrls (Anton 7.8.) 3.8. Remrk. Sometimes, integrls tht pper to be infinite in extent cn be given finite vlue in wy tht seems sensible. For exmple, consier x x If we rw picture of wht this might men grphiclly, in the sme wy s we i for integrls b f(x) x where n b re finite, we shoul be looking t the re of the region uner the grph y = x x region tht stretches infinitely fr into the istnce. So it seems infinite n nothing more to be si.

21 Techniques of integrtion But, before we conclue tht it is infinite, suppose we imgine colouring in the re uner tht grph with pint, n we o it so tht we pply the pint evenly so tht we use fixe mount per squre inch. The mount of pint we woul nee shoul be infinite if the re is infinite. We woul never be one pinting n infinite re, n so we coul pint wie but finite section n see how much pint we nee. In this picture, we show colouring for x 3, but we coul replce 3 by ny finite b >. The re covere up s fr s b works out s b [ x x = ] b = x b ( ) = b n we see tht, rther thn huge nswer, we lwys get n nswer <. In fct s b, the nswer pproches. It oes not ten to. So if we hve enough pint to pint squre unit of re, we will never completely run out lthough there will be very little pint left when b is lrge.mybe there is cse for eciing tht /x x shoul hve the vlue. We mke this our efinition Definition. We efine b x = lim x b x x (which is ). More generlly, if y = f(x) is efine n continuous for x we efine the improper integrl f(x) x = lim b b f(x) x if this limit exists n is finite. On the other hn if the limit oes not exist t ll, or is infinite, we sy tht the improper integrl f(x) x fils to converge.

22 05 6 Mthemtics MAE Exmple. Fin /x x. In this kin of exmple (n improper integrl) we strt by using the right efinition. This shows tht we relize tht there is n issue bout the integrl mking sense, n tht we know how the issue is elt with. We get /x x = lim b b /x x = lim [ln x] b b = lim ln b ln. b If we look t the grph of y = ln x we see tht this limit is. Becuse it is not finite, we sy tht the improper integrl x x oes not converge. Other types of improper integrl Integrls cn be improper becuse of verticl symptotes, or becuse the re uner the grph stretches infinitely fr wy in ifferent wys Definition. (i) If y = f(x) is efine n continuous for x b, then we efine the improper integrl b f(x) x = lim b f(x) x if this limit exists n is finite. If the limit oes not exist, or is infinite, we sy tht the improper integrl b f(x) x oes not converge. (ii) If y = f(x) is continuous on finite intervl < x b, excluing the left en point (where it might hve n symptote or other b behviour), then we efine the improper integrl b f(x) x = lim f(x) x c + c if this limit exists n is finite. If the limit oes not exist, or is infinite, we sy tht the improper integrl oes not converge. Exmple. Consier 0 /x x. Using the efinition 0 /x x = lim c 0 + = lim c 0 + b c [ x /x x ] = lim c c c

23 Techniques of integrtion 3 As this limit is infinite, the improper integrl 0 /x x oes not converge. (iii) If y = f(x) is continuous on finite intervl x < b, excluing the right en point b (where it might hve n symptote or other b behviour), hen we efine the improper integrl b f(x) x = lim c b c f(x) x if this limit exists n is finite. If the limit oes not exist, or is infinite, we sy tht the improper integrl oes not converge. (iv) If n integrl is improper for more thn one reson, or t plce not one of the enpoints, we ivie it s sum of improper integrls of the types we hve consiere bove. If ech one of the bits hs finite vlue, then the vlue of the whole is the sum. But if ny one of the bits fils to converge then the whole is si not to converge Exmples. () x x Here the problem with the integrl is t 0, where the integrn hs n symptote. In fct we lrey worke out tht 0 /x x oes not converge n so we know tht x lso x oes not converge. This is n exmple where n unthinking use of integrtion woul prouce wrong nswer. (You might be suspicious if the integrl of positive thing in the left to right irection turne out to be negtive.) (b) 0 + x x = + x x x x (It oes not mtter bout using 0 s stopping point. Any finite point woul o, for exmple x = 4 x + x) +x +x 4 +x We hve n ntierivtive tn x for +x n so we cn evlute the first integrl esily (c) 0 0 x = lim + x The other integrl 0 x = lim + x [tn x] 0 = x lso works out t π/. +x 0 lim (0 tn ( )) = π x( + x ) x = x( + x ) x+ x( + x ) x+ 0 x( + x ) x+ x( + x ) x The originl is improper becuse of the limit, the symptotes t x = 0 n the limit. Ech of the 4 bits bove hs just one problem t one en (n nees to be worke out s limit).

24 Mthemtics MAE0 3.9 Using computers to fin integrls We hve now conclue our reltively superficil stuy of methos of fining inefinite integrls. There is n rt to to being ble to use the methos well, but relly there re not so mny more such methos. However, s iscusse in Anton, there re some other pproches. In 7.6 Anton hs section clle Using computer lgebr systems n tbles of integrls. Tbles of integrls hve pretty much become obsolete now, but they re like the lists we hve in 3. bove or pge 6 of the formule & tbles prouce by the Stte Exmintions Commission only much longer lists. You use the tbles by trying to spot something in the list tht seems to help with your prticulr problem. There re vrious computer systems tht cn o this (or something more clever) for you. The most commercilly successful is Mthemtic (prouce by Wolfrm reserch) n it cn o most clcultions you cn o by hn (n more). There is website wolfrmlph.com which provies kin of Scientific version of Google serches n uses Mthemtic behin the scenes. It is perhps esier to use thn Mthemtic itself, s Mthemtic wnts you to follow its rules n complins if you on t ( bit like WileyPlus in tht respect). However wolfrmlph.com tries to give n nswer no mtter wht you sk it n so it cn be hr to control it to fin the specific nswer you wnt. Another computer lgebr system like Mthemtic is clle (Wterloo) Mple n it is lso commercil. There is free thing clle SgeMth n you cn ownlo it (it is very big) or use it online (see It probbly tkes bit of lerning before it woul be of use t ll. Wht is reltively new bout computer lgebr systems re tht () they cn compute with symbols, not just with numericl vlues n (b) moern computers such s your lptop re powerful enough for these systems. So they re in sense ccessible (though severl of them re expensive however see lso Computing with numbers is wht computers were oing for eces fter they were invente. They cn still o tht, your clcultor oes it on smll scle, n this sort of use is importnt for engineers who will now frequently simulte mchine numericlly before ttempting to buil it. One spect of such simultions often comes own to working out efinite integrls. How cn mchine clculte efinite integrl? In theory, this is esy. A efinite integrl b f(x) x is efine s limit of some finite sums clle Riemnn sums. Assuming n b re known exct numbers n f(x) is given by formul tht the computer cn evlute when fe vlue for x, the computer cn compute Riemnn sum (with lots of strips, or fine subivision of [, b]). Tht will give n pproximte vlue for b f(x) x. If the computer oes the sme clcultion more times with finer n finer prtitions, it cn get better n better pproximtions to the right nswer.

25 Techniques of integrtion 5 Refinements Now tht we hve stte tht it is possible to use the efinition irectly to compute efinite integrls b f(x) x numericlly (pproximtely), why o we nee to sy more? Well one question is bout efficiency. The metho might be refine to be more efficient n mybe lso to be ble to sy how pproximte the nswer is tht we get. There re questions like rouning errors tht cn buil up so tht longer clcultions cn become less ccurte if not one right. But there re lso cses where one wnts to estimte n integrl bse on smple vlues of the integrn obtine from n experiment or mesurement. In those cses, it my not be possible to keep gthering more t n it my be very esirble to use metho tht prouces the most ccurte nswer possible. There re two bsic methos of evluting efinite integrls b f(x) x numericlly, bse on vlue of the integrn t n + evenly spce vlues of x. If the spcing is h = (b )/n, then we suppose known (from mesurements or by clculting f(x)) the vlues of y = f(x) for x =, x = + h, x = + h,..., x = + nh = b We cn write x j = + jh for j = 0,,,..., n n y j = f(x j ) Trpezoil rule formul (Anton 7.7) b f(x) x = h (y 0 + y + y + + y n + y n ) Here is picture for = 0, b = n n = 4 ( very smll n). The formul clcultes the integrl of the function with grph me of bits of stright lines linking the points (x j, y j ) on the grph y = f(x). It shoul be better estimte of the integrl tht squring off the verticl strips s one oes for Riemnn sums.

26 Mthemtics MAE Simpson s rule formul (Anton 7.7) Here the nottion is s before but n hs to be even. b ( f(x) x = h 3 y y + 3 y y y n y n + ) 3 y n The ie behin this is to replce the ctul grph y = f(x) by grph me up of bits of qurtic grphs. The first qurtic grph goes through (x 0, y 0 ), (x, y ) n (x, y ). The next strts t (x, y ) n goes through (x 3, y 3 ) n ens t (x 4, y 4 ) n so on. Just s there is one line through points, there is one qurtic through 3 points (with ifferent x coorintes) n the formul comes from working out wht tht grph is n then its integrl. There re theorems (given in Anton) explining theoreticl worst-cse error estimtes from using the Trpezoil rule or Simpson s rule. More or less, Simpson s rule is likely to be more ccurte thn the Trpezoil rule, or to give gurntee ccurcy with smller n. Somewht more ccurtely, the worst possible error in the Trpezoil rule is proportionl to h n in Simpson s rule proportionl to h 4. If h is firly smll (which mens n big since h = (b )/n), h 4 will be lot smller thn h. However the constnts of proportionlity cn be ifferent, so tht the e ccurcy in Simpson s rule might not kick in for moest vlues of n you use. So if you wnt to clculte ln(.0) = (/x) x n you wnt to be sure tht the nswer you get is goo to (sy) 4 eciml plces, then you cn use these theorem to fin n n so tht the Trpezoil rule (or Simpson s rule) with tht n will surely be tht ccurte. This ie of worst-cse nlysis my be unuly cutious in some cses, n in prctice the methos cn often be use without being sure of the errors. In cse where you only hve t, sy = 0 + eqully spce vlues of the integrn, these worst cse results won t be usble becuse they epen on knowing f(x) n its erivtives. The methos cn still be use s estimtes, but the errors will nee to be mnge in some other wy. Richr M. Timoney Februry 7, 06

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