Techniques of Integration

Transcription

1 Chpter 8 Techniques of Integrtion 8. Integrtion by Prts Some Exmples of Integrtion Exmple 8... Use π/4 +cos4x. cos θ = +cosθ. Exmple 8... Find secx. The ide is to multiply secx+tnx both the numertor nd denomintor: secx = secx secx+tnx secx+tnx sec x+secxtnx = secx+tnx du = u = ln secx+tnx +C. Similrly, we obtin cscx = ln cscx+cotx +C.

2 CHAPTER 8. TECHNIQUES OF INTEGRATION Integrl tbles () +u du = u tn () u du = u sin ( > ). ( > ). Exmple For /(4 + 9x ), use substitution first. Let 3x/ = u then 3/ = du, nd 4+9x = 4 = 6 +( 3x ) +u du = 6 tn 3 x+c. Integrl by prts Integrting w.r.t x d (uv) = udv +vdu. uv = = u dv + v du udv + vdu. Thus Proposition 8..4 (Integrtion by Prts I). udv = uv vdu. (8.) Proposition 8..5 (Integrtion by Prts II). f(x)g (x) = f(x)g(x) f (x)g(x). (8.) Proposition 8..6 (Definite integrl). b b f(x)g (x) = [f(x)g(x)] b f (x)g(x).

3 8.. INTEGRATION BY PARTS 3 Exmple Find the following π () () xsinx lnx. sol. () Let u = x, dv = sinx. Then du =, v = cosx. (Fig 8.) π π xsinx = [x( cosx)] π ( cosx) = π +[sinx] π = π. () Let u = lnx, dv =. Then we hve du = (/x), v = x. lnx = (lnx)x = xlnx x+c. x x y y = xsinx π x Figure 8.: Repeted integrtion by prts Exmple Find x sinx. sol. Let u = x, dv = sinx. Then du = x, v = cosx nd hence x sinx = x ( cosx) ( cosx)x = x cosx+ xcosx.

4 4 CHAPTER 8. TECHNIQUES OF INTEGRATION f nd its derivtive g nd its integrl x (+) e x x ( ) e x (+) e x e x Agin, set u = x, dv = cosx. Then du =, v = sinx. x sinx = x cosx+xsinx sinx = x cosx+xsinx+cosx+c. Exmple Find x e x. sol. f(x) = x, g(x) = e x f nd its derivtive g nd its integrl x 3 (+) sinx 3x ( ) cosx 6x (+) sinx 6 ( ) cos x sinx Exmple 8... Find x 3 sinx. Use the tble bove x 3 sinx = x 3 cosx+3x sinx+6xcosx 6sinx+C. Exmple 8... Find e x sinx. sol. If u = e x, dv = sinx, then du = e x, v = cosx. e x sinx = e x ( cosx) e x ( cosx) = e x cosx+ e x cosx.

5 8.. INTEGRATION BY PARTS 5 Agin let u = e x, dv = cosx so tht du = e x, v = sinx. e x sinx = e x cosx+ e x cosx = e x cosx+e x sinx e x sinx. Solving this for e x sinx we obtin e x sinx = ex (sinx cosx)+c. Reduction formul Exmple 8... Express cos n x in terms of low power of cosx. sol. cos n xcosx = cos n sinx+(n ) = cos n sinx+(n ) = cos n sinx+(n ) sin xcos n x ( cos x)cos n x cos n x (n ) cos n x. So n cos n x = cos n sinx+(n ) cos n x. Exmple Prove ( ±x ) n = x( ±x ) n+ + n n+ ( ±x ) n, (n ).

6 6 CHAPTER 8. TECHNIQUES OF INTEGRATION sol. Integrtion by prts ( ±x ) n = x( ±x ) n x n( ±x ) n (±x) = x( ±x ) n n( ±x ) n ( ±x ) = x( ±x ) n n ( ±x ) n +n ( ±x ) n. If n /, ( ±x ) n = x( ±x ) n n+ + n n+ ( ±x ) n. 8. Integrtion of Trigonometric function Products of powers of Sines nd Cosines Integrl of sin m xcos n x () If m is odd, then set m = k + nd use sin x = cos x sinx = d(cosx) to trnsform it to sin k+ xcos n x = ( cos x) k cos n xd(cosx). () If n is odd n = k +, use cos x = sin x cosx = d(sinx) to obtin sin m xcos k+ x = sin m x( sin x) k d(sinx). (3) If both m,n re even, use sin x = ( cosx)/, cos x = (+cosx)/ to lower the degree nd repet the previous technique. Exmple 8... Find sol. sin 5 x = sin 5 x. ( cos x) d(cosx)

7 8.. INTEGRATION OF TRIGONOMETRIC FUNCTION 7 = ( cos x+cos 4 x)d(cosx) = 5 cos5 x+ 3 cos3 x cosx+c. Exmple 8... Find sin xcos 3 x. sol. sin xcos 3 x = sin x( sin x)d(sinx) = 5 sin5 x+ 3 sin3 x+c. Exmple Find sin 4 xcos x. sol. ( ) cosx ( ) +cosx sin 4 xcos x = = 8 ( cosx+cos x ) (+cosx) = ( cosx cos x+cos 3 x ) 8 = ( cosx +cos4x ) +( sin x)cosx 8 = ( cos4x sin x cosx ) 6 = ( x 6 4 sin4x ) 3 sin3 x +C. Integrl of ±sinx, ±cosx Use the double ngle formul. sina = sinacosa cosa = cos A = sin A. Chnge the form ±sinx, ±cosx to complete squre. Exmple Find π sinx.

8 8 CHAPTER 8. TECHNIQUES OF INTEGRATION sol. Use the identity: sinx = sin( x )cos(x ) = (sin( x ) cos(x ) ). π π sinx = sin x cos x π/ ( = cos x sin x ) π + = [ sin x +cos x ] π/ + π/ ( sin x cos x ) [ cos x sin x = ( + )+( + + ) = 4( ). ] π π/ Exmple Find π/ sol. +cosx = cos x, π/ +cosx. π/ +cosx = cosx = [sinx] π/ =. Tngent nd secnt Recll Exmple tn x = sec x, (tnx) = sec x, (secx) = secxtnx. secx.

9 8.. INTEGRATION OF TRIGONOMETRIC FUNCTION 9 sol. Multiply secx+tnx. secx(secx+tnx) secx = secx+tnx (secx+tnx) = secx+tnx = ln secx+tnx +C. Exmple tn xsecx. sol. Since tn xsecx = (sec x )secx = sec 3 x secx, we cn find sec 3 x. Let u = secx, dv = sec x then v = tnx, du = secxtnx, we hve sec 3 x = secxtnx = secxtnx = secxtnx (tnx)secxtnx (sec x )secx sec 3 x+ secx. Hence we obtin sec 3 x = secxtnx+ secx. Hence tn xsecx = sec 3 x secx = secxtnx secx = secxtnx ln secx+tnx +C. Exmple tn 6 x.

10 CHAPTER 8. TECHNIQUES OF INTEGRATION sol. Since tn x = sec x tn 6 x = tn 4 x(sec x ) = tn 4 xsec x tn 4 x = tn 4 xsec x tn x(sec x ) = tn 4 xsec x tn xsec x+ tn x = tn 4 xsec x tn xsec x+ (sec x ) = 5 tn5 x 3 tn3 x+tnx x+c. Remrk For cotx or cscx, use +cot x = csc x, (cotx) = csc x, (cscx) = cscxcotx. Products such s sinmxsinnx, sinmxcosnx, cosmxcosnx Addition formul: sin(a+b) = sinacosb +cosasinb sin(a B) = sinacosb cosasinb cos(a+b) = cosacosb sinasinb cos(a B) = cosacosb +sinasinb. From these we get(with A = mx, B = nx) sinmxsinnx = [cos(m n)x cos(m+n)x] sinmxcosnx = [sin(m n)x+sin(m+n)x] cosmxcosnx = [cos(m n)x+cos(m+n)x].

11 8.3. TRIG SUBSTITUTION Exmple 8... π/6 sin4xsin3x. sol. π/6 π/6 sin4xsin3x = (cosx cos7x) = [sinx 7 ] π/6 sin7x = Trig Substitution Qudrtic term For the terms of the forms u, +u u, we cn try to substitute u = sinθ, u = tnθ, u = secθ resp. u = sin θ = ( sin θ) = cos θ (8.3) +u = + tn θ = (+tn θ) = sec θ (8.4) u = sec θ = (sec θ ) = tn θ. (8.5) Note the domin of definition () u = sinθ is defined on π/ θ π/. () u = tnθ θ = tn (u/) on π/ < θ < π/. (3) u = secθ θ = sec (u/) Since u θ < π/ (if u ), or π/ < θ π (if u ). Exmple du +u.

12 CHAPTER 8. TECHNIQUES OF INTEGRATION +x x x x x x x = tnθ x = sinθ x = secθ Figure 8.: trig substitution sol. Use substitution u = tnθ, du = sec θdθ to get du sec +u = θdθ sec θ dθ = = θ +C = tn u +C. Exmple Find u du, ( > ). sol. Use u = sinθ, du = cosθdθ to get u du = = = = cosθ cosθdθ (+cosθ)dθ ( θ + sinθ )+C = (θ +sinθcosθ)+c ( sin u + u ) u +C = sin u + u u +C.

13 8.3. TRIG SUBSTITUTION 3 Exmple Find du ( u > > ). u, sol. Let u = secθ u = (sec θ ) = tn θ, du = secθtnθdθ. Then du secθtnθdθ u = tnθ = { secθdθ ( < θ < π/) secθdθ (π/ < θ < π) ln secθ+tnθ +C ( < θ < π/) = ln secθ+tnθ +C (π/ < θ < π) u u ln + +C (u > ) = u u ln +C (u < ). Lst integrls cn be simplified s follows: u u ln + u+ ln. = ln u u u ln = ln u u (u+ u = ln ) (u u )(u+ u ) (u+ u = ln ) = ln u+ u = ln u+ ln. u

14 4 CHAPTER 8. TECHNIQUES OF INTEGRATION Hence du u = ln u+ u +C. Exmple x +9. sol. Let x = 3tnθ ( π/ < θ < π/), = 3sec θdθ, 3sec x +9 = θ 3secθ dθ = secθdθ = ln secθ+tnθ +C (x ) x = ln C = ln x+ x +9 +C. Involving x +bx+c Completing the squre For fctors like x +bx+c, (,b ), use u = x+b/() to get x +bx+c = (u ±p ). x x Exmple Find. sol. Since x x = (x ) u = x we hve s in exmple 8.3. with =, x x u = du = sin u+ u u +C = sin (x )+ (x ) x x +C.

15 8.4. INTEGRATION OF RATIONAL FUNCTIONS 5 Exmple x +x+. sol. x +x+ = (x+/) +3/4 u = x+/ = 3/ x +x+ = du u +3/4 = u tn +C 3 3 = x+ tn +C Integrtion of Rtionl functions When p(x), q(x) re rtionl functions, we cn lwys write it s p(x) r(x) = Q(x)+ q(x) q(x) for some polynomil Q(x),r(x) (degree of r(x) is less thn tht of q(x).) Distinct liner fctors Suppose α,...,α r re distinct nd p(x) is polynomil of degree of is less thn r. Then we cn set p(x) (x α ) (x α r ) = A x α + + A r x α r. (8.6) Here A i s cn be obtined by method of undetermined coefficients.(there is nother method, clled Heviside cover up method, see below) Exmple Find r (x α ) (x α r ) = A i ln x α i +C. x+ x(x+). i=

16 6 CHAPTER 8. TECHNIQUES OF INTEGRATION sol. One cn find the following prtil frction x+ x(x+) = x + (x+). x+ x(x+) = ( x + ) x+ = ln x(x+) +C. Exmple Find x+ x 3 x. sol. Since x 3 x = x(x )(x+) we cn set x+ x 3 x = A x + B x + C x+. Solving for A,B,C we get A =, B = 3/, C = /. Hence ( x+ x 3 x = x + 3/ x + / ) x+ = ln x + 3 ln x ln x+ +C. Repeted liner fctor Assume the degree of p(x) is less thn tht of r(x). Then p(x) (x α) r = A x α + A (x α) + + A r (x α) r. To find the coefficients A,A,...,A r, multiply (x α) r. Then p(x) = A (x α) r +A (x α) r + +A r.

17 8.4. INTEGRATION OF RATIONAL FUNCTIONS 7 Now use method of undetermined coefficients to find A i s. Another nice wy of finding A i s by derivtive will be introduced below. Once A is re known, we cn find the integrl: ( ) p(x) (x α) r = A x α + A (x α) + + A r (x α) r Exmple Find = A ln x α A x α (r )A r (x α) x (x ) 3. sol. Since x = (x ) +4(x )+4, we hve x (x ) 3 = x + 4 (x ) + 4 (x ) 3. r +C. Hence x ( ) (x ) 3 = x + 4 (x ) + 4 (x ) 3 = ln x 4 x 8 (x ) +C. Irreducible qudrtic fctor Suppose x +β x +γ,...,x + β r x +γ r re distinct qudrtic fctor without hving rel roots(we sy irreducible qudrtic fctor). Suppose p(x) is polynomil of degree less thn r. So we hve p(x) r (x +β x+γ ) (x +β r x+γ r ) = i= B i x+c i x +β i x+γ i for some B,...,B r nd C,...,C r. Hence p(x) r (x +β x+γ ) (x +β r x+γ r ) = i= B i x+c i x +β i x+γ i.

18 8 CHAPTER 8. TECHNIQUES OF INTEGRATION Agin we cn find the coefficients by method of undetermined coefficients. Now since we hve B i x+c i = B i (x+β i)+d i, (D i = C i B i β i /) = B i (x +β i x+γ i ) +D i, ( B i x+c i Bi (x +β i x+γ i ) x = +β i x+γ i x + +β i x+γ i = B i ln(x +β i x+γ i )+ For D i /(x +β i x+γ i ) use the formul: Exmple Find du u + = tn u +C. x x 4 +x +. sol. Since x 4 +x + = (x x+)(x +x+), we set x x 4 +x + = B x+c x x+ + B x+c x +x+. ) D i x +β i x+γ i D i x. +β i x+γ i By compring, we obtin B = B =, C =, C =. Since x ±x+ = (x±/) +( 3/), we see x x 4 +x + ( = (x /) +( 3/) (x+/) +( 3/) = ( x x+ tn tn )+C )

19 8.4. INTEGRATION OF RATIONAL FUNCTIONS 9 Repeted irreducible qudrtic fctor Suppose p(x) is polynomil of degree less thn r, nd x +βx+γ does not hve rel roots. Then we cn set p(x) (x +βx+γ) r = B x+c x +βx+γ + B x+c (x +βx+γ) + + B rx+c r (x +βx+γ) r for some B,B,...,B r,c,c,...,c r. Then p(x) (x +βx+γ) r ( B x+c = x +βx+γ + B x+c (x +βx+γ) + + B ) rx+c r (x +βx+γ) r. By the sme wy s before we see, with D i = C i B i β/ ( B i x+c i (x +βx+γ) i = Bi (x +βx+γ) ) (x +βx+γ) i + D i (x +βx+γ) i B i = (i )(x +βx+γ) i + D i (x +βx+γ) i. For the integrl of D i /(x +βx+γ) i (i ), use the recurrence reltion du (u + ) i = u i 3 (i )(u + + ) i (i ) du (u + ) i. Exmple Find x 4 +x 3 +5x +6 (x +) 3. sol. x 4 +x 3 +5x +6 (x +) 3 = A x+b x + A x+b + (x +) + A 3x+B 3 (x +) 3. Multiply (x +) 3 to see x 4 +x 3 +5x +6 = A x 5 +B x 4 +(4A +A )x 3 +(4B +B )x +(4A +A +A 3 )x+4b +B +B 3. Compring, we get A =, A =, A 3 =, B =, B =, B 3 =. Hence

20 CHAPTER 8. TECHNIQUES OF INTEGRATION the integrnd is Hence x 4 +x 3 +5x +6 (x +) 3 = x + + x+ (x +) + 4x (x +) 3. x 4 +x 3 +5x +6 (x +) 3 = x + + x (x +) + = tn x x + + (x +) + x 4(x +) + 4 = 5 4 tn x + x 4 4(x +) + (x +) +C. 4x (x +) 3 x + + (x +) Heviside cover up method for liner fctors Exmple x + (x )(x )(x 3) = A x + B x + C x 3. Here A = () + (x-) cover ( )( 3) B = C = () + ( ) (x ) ( 3) = 5 ()( ) = 5 cover (3) + = (3 )(3 ) (x 3) ()() = 5. cover Exmple Do the sme with x+4 x(x )(x+5).

21 8.5. INTEGRAL TABLES AND CAS sol. Note x+4 x(x )(x+5) A = B = C = = A x + B x + C x+5 +4 x ( )(+5) = 5 +4 (x ) (+5) = ( 5)( 5 ) (x+5) = 35. Using differentition-repeted fctors Exmple x (x+) 3 = A x+ + B (x+) + C (x+) 3. Write x = A(x+) +B(x+)+C. Substitute x = to get C =. Then tke derivtive = A(x+)+B nd substitute x = to get B =. Finlly, tking derivtive gin, we see A =. 8.5 Integrl Tbles nd CAS Exmple Find xsin x. sol. We use the formul (derive it?) x n sin x = xn+ n+ sin x n+ x n+. x,n

22 CHAPTER 8. TECHNIQUES OF INTEGRATION Integrtion with Mple For the indefinite integrl of f(x) = x +x in Mple, type > f:=x^*sqrt(^ +x^) > int(f,x) Then you get the nswer. 8.6 Numericl Integrtion Trpezoidl Rule To evlute the definite integrl b f(x) we divide the intervl by n (uniform) subintervl nd set = x < x < < x n = b. Here x = h = b n nd x i = +(i ) x,i =,,n. With y xi = f(x i ) we use trpezoidl rule on ech subintervl to get b f(x) h (y +y + +y n +y n ). (Error) = E T M(b )3 n. h h Figure 8.3: Trpezoidl Rule

23 8.6. NUMERICAL INTEGRATION 3 Simpson s Rule Replce the definite integrl by n integrl of qudrtic interpoltion. Exct for poly. of degree three. Assume y = Ax + Bx + C is n interpolting polynomil of f in the sense tht y(x i ) = f(x i ) for x = h,x =,x = h h h f(x) h h (Ax +Bx+C) = Ax3 3 + Bx +Cx = Ah3 3 ] h h +Ch = h 3 (Ah +6C). Since y = Ah Bh+C, y = C, y = Ah +Bh+C we see nd the the integrl is A = y y +y h, B = y y h, C = y h 3 (y y +y +6y ) = h 3 (y +4y +y ). Since this formul is exct for x 3, it is in generl third order formul. When h h Figure 8.4: Simpson s Rule the generl intervl [,b] is divided by n even number of intervls, we cn pply it repetedly to get b f(x) h 3 (y +4y +y +4y 3 + +y n +4y n +y n ).

24 4 CHAPTER 8. TECHNIQUES OF INTEGRATION E S M(b )5 8n 4. Exmple Find n upper bound for the error in estimting 5x4 using Simpson s rule with n = 5. sol. Let f(x) = 5x 4. Then f (4) =. So M =. b = nd n = 4. The error bound is E S M(b )5 8n 4 = () =. Exmple Wht is the minimum number of intervls needed to pproximte bove exmple using the Simpson s rule with n error less thn 4. sol. We set Then () 5 8n 4 M(b ) 5 8n 4 < 4. < 4 n 4 > 64()4 3 ( ) 64 /4 n > Improper Integrl So fr the integrl ws defined only when () The domin is finite like [,b]. () The rnge of the function is finite In prctice, there re cses when either one or both of these conditions violtes. Improper Integrl Definition 8.7. (Improper integrl). () When = or b =,

25 8.7. IMPROPER INTEGRAL 5 () or f is undefined (infinite vlue) t either or b the integrl b The Cse when or b is f(x) is clled n improper integrl. Definition 8.7. (Convergence of Improper integrl). () Suppose f(x) is continuous on [, ). We set provided the limit exists. f(x) = lim b b () Similrly, if f(x) is continuous on (,b], we set b provided the limit exists. f(x) = lim b (3) If f(x) is continuous on (, ) then we set b f(x) = lim b f(x) (8.7) f(x) (8.8) f(x) (8.9) provided the limit exists. In these cses, we sy the improper integrl converges. Otherwise, we sy the integrl diverges. Exmple Exmple ln x +x The Cse when f is undefined (infinite vlue) t either or b Definition (Convergence of Improper integrl). () Suppose f(x) is integrble on ll closed subintervl of [,b) nd we hve either lim = ±. If the limit x b f(x) L = lim u b u f(x) (8.)

26 6 CHAPTER 8. TECHNIQUES OF INTEGRATION exists then we sy the improper integrl converges nd write its limit b f(x) = lim u b u f(x). () The sme definition holds when lim = ±. We write x +f(x) b f(x) = lim l + b l f(x) (8.) if the ltter limit exists. Otherwise, we sy the integrl diverges. (3) The discontinuity cn hppen t n interior point. In this cse, we cn still pply the bove definitions. Computtion of Improper integrl y y y = f(x) y = f(x) u b x u x Figure 8.5: Improper integrl on [, b) y y y = f(x) y = f(x) l b x l b x Figure 8.6: Improper integrl on (, b] Exmple Find the re surrounded by y = / x, x-xis, y-xis, x = (fig 8.7). sol. The function / x is not defined t x =. But we cn use the limit

27 8.7. IMPROPER INTEGRAL 7 y y = x x Figure 8.7: Improper Integrl s (Are) = lim ε + ε x [ = lim x /] ε + ε ( = lim ε /) =. ε + Exmple x. sol. We distinguish two cse: (,] nd [,). = lim x l + l x [ = lim sin x ] l + l = sin ( ) = π. u = lim x u x [ = lim sin x ] u u = sin () = π. Hence = x + x = π. x

28 8 CHAPTER 8. TECHNIQUES OF INTEGRATION Exmple (x ) 4/3. sol. The function /(x ) 4/3 is not defined t x =. Hence we seprte (x ) 4/3 = (x ) 4/3 + (x ) 4/3. Since u = lim (x ) 4/3 u (x ) 4/3 [ = lim 3(x ) /3] u u 3 = lim u 3 (u ) /3 =. divergestheintegrldivergesregrdlessof (x ) 4/3 (x ) 4/3. y y = (x ) 4 3 x Figure 8.8: The function /x p Theintegrl of /x p on (,] or [, ) dependson the vlueof p. Inprticulr, the integrl on [, ) is used to decide the convergence of the series /n p.

29 8.7. IMPROPER INTEGRAL 9 On (,] Exmple Find (p > ). xp y x Figure 8.9: On (,] sol. () For < p < () For p = [ x p = lim x p ] l + l x p = lim l p = lim l + p l l + p (3) For p > = p. u x p = lim l + x = lim l +[lnx] l = lim =. l +( lnl) [ ] x p = lim x p l + l x p = lim l p = lim l + p l l + p =. On [, ) Exmple Find sol. () For < p <, (p > ). xp u [ ] x p = lim x p u u x p = lim u p = lim =. u p u p

30 3 CHAPTER 8. TECHNIQUES OF INTEGRATION y x Figure 8.: Improper integrl on [, ) () For p = (3) For p > u x p = lim u x = lim u [lnx]u = lim lnu =. u u [ x p = lim x p ] u u x p = lim u p = lim = u p u p p. Test for Convergence Theorem 8.7. (Comprison test). Let f(x) g(x) for ll x >. Then () If () If g(x) converges, then f(x) diverges, then Exmple Test whether f(x) lso converges. g(x) lso diverges. converges or not? +x3 sol. We see, for ll x, /(+x 3 ) /x 3 holds. By exmple 8.7. we see /x 3 = /. Hence by Comprison test /(+x 3 ) converges. On the other hnd, the integr /(+x3 ) is well defined on [,]. Hence /(+x 3 ) converges nd the vlue is /(+x3 )+ /(+x 3 ). (See Fig 8.)

31 8.7. IMPROPER INTEGRAL 3 y y = +x 3 y = x 3 y = +x x Figure 8.: Theorem (Limit Comprison Test). Assume f(x), g(x) re positive on [, ) nd suppose f(x) lim = L( < L < ). x g(x) Then the two integr g(x) both converge or both diverge. f(x) nd Proof. () Suppose g(x) converges: Then there is N > such tht f(x)/g(x) L+holdsforllx N. Sowehve f(x) (L+)g(x) nd by Limit Comprison Test, N f(x) converge. Hence f(x) converges to N f(x)+ N f(x). () Suppose g(x) diverges:there exists N > s.t. for ll x N, f(x)/g(x) L L/ = L/ holds. Hence f(x) (L/)g(x) nd by Limit Comprison Test N f(x) diverges. So does f(x). Exmple Test whether sol. Let f(x) = /(+e x ), g(x) = /e x. Then nd converges or not? +ex f(x) lim x g(x) = lim e x x +e x = u e x = lim u e x = lim [ e x ] u u = lim ( e u + ) =. u

32 3 CHAPTER 8. TECHNIQUES OF INTEGRATION Hence by Limit Comprison Test, /(+e x ) converges. Exmple Test for convergence sol. Set f(x) = Then x x nd g(x) = x. x x. f(x) x lim x g(x) = lim x x =. By Limit Comprison Test, [ ] ( u = lim x x u = lim u ) =. u x x converges. y y = x x x Figure 8.: