University:
California State University, Northridge
Course:
MATH 150B | Calculus II
Academic year:

2023
Views:

161

Pages:

5
Author:

Yung D

8.7 Improper Integrals (Type I and Type II) Def: Improper Integrals of Type I infinite limits of integration: are integrals with 43 00 f(x)dx = lim a b Sfix)dx where f is continous a a on [a,00) b00 12 If f is continuous on (-0, a), then Sfondx = lim (fox)dx -00 BJ If f is continueous f(x)dx -00 on (-10,00), then. C de dx In each case, if the limit is finite, then the improper integral converges and it is equal to this limit Area" Other wise the improper integral divergesiait Ex dx x²+1 b 6 = lim = lim tank != tai'b= de x²+1 lim -2 2 dx -2 2 dx (x-1) (x+1) 2 (x-1)(x+1) 1-8 A = = -00 - xp( x+1 = lim - x-1 x+1 9 A=1 B=-1 -2 = lim -2 X +1 b b = lim (44) btl b-00 = In3-In limb! b- b+1 = In3-Inl In 3 Note that de 00 1+x² ==+=ㅠ dx = + = 1+x² 1+x2 EXP b 16 tea's dr = lim 16 tarik de tanx dx 1+ x2 tan b tanx dx 1+ x = lim | 16 udu = lim su 2 = lim 8 [(tan'b)² - (tan'o)] = $()² = 2ㅠ² 2 tan b -1 b00 u=tan's du dx Def Improper Integrals of Type II are integrals of functions that become infinite at a point within the interral of integration Vertical Asymptoles) b III If f is discontinuous at a, then (food x = lim (f(x)dx a 2 If f is discontinuous at b, then f(x)dx 1x = lim cob b 13) If f is discontinuous at c, where acrab, then In each case, if the limit is finite Sfoodx a C Sfoode=ffoods a then the improper integral converges and it is equal to this limit "Area". Otherwise the improper integral diverges infinite area" b Tests for Convergence and Divergence (46) Th(Direct Comparasion Test) Let f and g be continuous on [a, 00) with os f(x) = g(x) for all xa. 00 * If I g(x) dx converges, then Sfindx converges. a a * If $f(x)dx diverges, then {g(x)dx diverges. a 20 10 {(x) 9Cx) sinx on [1,0) dx diverges because 00 Ensin dx Exp sin's dx converges because 2 and Convergs see exp*" Thus, by the Direct Comparasion Test (sin-dx converges 10 [1,00) ٥٠١ 00 and diverges "see exp *" Thus, by the Direct Comparision Test, usta, Singer dx 00 [0,1] Edisint converges because of tron [o] ≤ + +sint √t and = lim 2 VE bo lim bo = 2√π which converges, Thus, by the Direct Comparasion test, sdt Ietsint Converges. b = 900-9 Th (limit Comparision Test) (47) If f and gare positive and continous functions on [a, 0) lim f(x) x→00 g(x) = L where <<8, ge or both diverge. then Sfoxidx and Socx)dx both converge Exp dx 1+ x • let f(x) = f and g Are g(x)= positive continuous 1 1+x2 on ]1,ه lim f = lim = lim 1tx = 1 "finite+" and dx Converges "see exp". Thus, by the Limit Comparesion Test da converges. 2 √x-1 00 Let f(x) = 9 (x) = f and g are positive continuous on [2,00) lim f = lim 9 lim lim 1 and = lim 2 b b • Sx & dx = lim 2√x = lim [206-22] = +00 2 diverges by the limit Comparasion Tast 2 which diverges > 2 VX-1 00