Lecture Note
University
California State UniversityCourse
MATH 150B | Calculus IIPages
5
Academic year
2023
yung dump
Views
0
8.7 Improper Integrals (Type I and Type I 43 Def : Improper Integrals of Type I are integrals with infinite limits of integration: [I] Sixdx = lim Sfowdr b where f is continous on [a,00) a b 8 a 2 If f is continuous on (-00, a), / then a a S f(x) d x = lim f(x) dx b - 00 - 00 3) If f is continues on (-00,00), then 00 as C / f(x) d x = S'fusde + fusd , CEIR -00 -00 c In each case, if the limit is finite, then the improper in tegral / converges and it is equal to this limit? Aren'' II 1/11 HE O ther wise the improper integral diverges 16 Area infinit EXOS . dx = lim S dx = lim tanx I' = lim tan b = HI X + x2+1 b-soo b-300 O 2 5 x2-1 2 dx = (x-1) 2 (x+1) dx (x-1)(x+1) 2 = x-1 A + x+ B \ B=-1 A=1 -00 - 00 2 = I I -) x+1 - & -2 = lim ) dx - b 8 x
-2 = lim In /x-1/ - b-> 8 mlx+11) 44 b -2 = lim b -00 [In I = lim [1]] -3/ - 1n/b-11 b+1 b b-> w = In3- In lim b-1 = In3- - In/ = In3 b .00 b+1 that 5" dx dx = dx Note + = II + 2 IA = 1+x2 1+x2 -00 -00 o b ustan's Exp s 16 tan dx = lim 16 tan dx 1+x2 b It ? du= dx 8 X O 1+x2 tan b 0 tan'b = lim s 16 u du = lim 8u2 b 00 b-s 00 0 O = lim 8 [(tan'b) - (tan'o) ] = 8 () = 2 II b-> 8 Def Improper Integrals of Type II are integrals of internal functions that become infinite at a point within the II If f is discontinuous at a, then Sfudd== = lim Sfredd of integration(Vertical Asymptotes) a c->at C C 2 If f is discontinuous at b, then Sfwaxx = lim foods a b a b [3) If f is discontinuous ate, where acceb, then Sfoodx b = Sfwaxx C ffoods a In each case, if the limit is finite, / then the improper a integral converges and it is equal to this limit Area". otherwise the improper integral diverges 8 infinite area"
Exp S VX dx = lim o+ S' dr VX = (-)0" lim 2vx! c Hi 45 c-> c O = lim (2 - 2VE) = 2- 0 = 2 2 c-so+ b 4 Exp S" dx = lim dx = lim -254-x V 4- - b-s 4- b-4 o X o 0 = lim [-204 -b +4]=4 + = b->4 as Exp For what values of P does the integral S dx XP converge? D C 1-P c P#1 S dx = lim x-dx = lim X I xp C & C 8 1-P = lim [ c - c- 8 1-P 1- P { , p>l = / ps X I b-so 2 8 Exp 5" dx S dx to sec'x dx = + / xvx-1 I X x2-1 2 C = lim I' dx lim S dx I 2 + x xx - 1 xvx2-1 -2 bost b 8 2 2 C = b-s/+ lim [see 1x1) + lim [sec'1*1] c b = lim b->1+ [setz- sec'b] + lim 00 [see'c-spcr] - The 3 = lim - sec b + lim sec's b->1+ C->00 O + II 2 = II
Tests for Convergence and Divergence 46 Th ( Direct Comparasion Test) Let f and g be continuous on [a, o) with os f(x) g(x) for all X a. 00 & If S g(x) dx converges / then Sfixdx converges. a a If S 00 f(x) dx diverges, then Sgx)dx diverges. a a ([x) g(x) sin 2 8
Th ( limit Comparision Test) 47 If f and g are positive and continous functions on [a,as) lim F(x) = L where OLL< 8, x->00 g(x) 8 & then fuxixx and Sgxxdx both converge or both diverge. a a \ 8 Ex 5 dx let f(x) = 1/2 / g(x) \ + x2 I It x2 f and g are positive continuers on [1,00 lim f = lim 1/x = lim 1+x2 = 1 "finite+" the ->00 g X-00 1 x & 1+x2 and dx converges "see exp*" Thus, by the Limit Comparasion Test x2 & dx converges. It x2 II 4 I Exp dx . Let frx) = , g(x) = r X X-1 X-1 2 f and g are positive continuous on [2, os) lim f = lim VX-1 = lim Tx = lim 1 = 1 x 00 g x-so NO 00 V X-1 X-DOO VI-1 8 rx and S dx = lim 2 dx = lim 2VX = lim [2UT - 2VE] = + 8 Vx 2 b-so b-> & 2 which diverges D dx diverges by the limit Comparasion Test. 2 Vx-1
Improper Integrals
Please or to post comments