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# 05.06 - 5.6 SUBSTITUTION AND AREA BETWEEN CURVES 4(a...

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Unformatted text preview: 5.6 SUBSTITUTION AND AREA BETWEEN CURVES 4. (a) Letuzcosx 2:Β» duzβsinxdx :> βdu:sinxdx;xβ0 β> uβ1,x_1r _> u_.β1 7r β] Β£3cos2xsinxdx= I: β-3I.i2'du=1βu3_1;1 =β(β1)3β(β(1)3)=2 (b) Use .the same substitution asinpart (a);x 22% 2> u21-,X2 37; 23> u 2 β1 37:β *1 I; 3cos2xsinxdx=c β31;72 du'=2 6. (a) Letu=t2+1Β’ du'=2tdt => %'du=_tdt;t=0 => u=1,t=\ο¬-=> 11:8 {-7 3 ' . C W +1>β3d J, w an m {(4).(2)u?4/31~β~%( )(8>4/3β(%)(1)β/β2 ββ (b) Use the same substitution asinpart (a); t : β\/_ => .u = 8, t: 0 => 11 = 1 0 i I 8 ff t(t2+1)1/3dt =Β£%u1/3du=β' I; Γ©ul/sdu-=β% 30. Letu2lnx 25 dU2idx;X22 2β,Β» u21nZandX24 2:Β» 1121114; 4 "In I dx = f 4ldu=1111311β::ln(ln4)βln(ln2)=1n1L; ):In(Lβ-1;1222)β 2ββ mil =ln2 58. We want the area between the x-axis and the curve yβ = 2 ,0 < x < 1 plus the area of a triangle (formed by x = 1, -x+y2 2, .and the xβaxis) with base 1 and height]. Thus, A2 ο¬x? dx+β 2'(1)(1)2 1%31Z-lβ212315-l- 1β 59. AREA =-A1 + A2 A1: For the sketch given, a = β3 and we find b by solving the equations y = x2 β 4 and y = -x2 β 2x simultaneomly for x: x2 β 4 = βx2 β 2x => 2x2'+ 2x β 4 = 0 => 2(x + 2)(x - 1') => x = β2 or x: 1 so b2 β2: f(x) β g(x) 2 (x2 β4) β (βx2 β 2x) 2 2x2 + 2x β 4 2> A1 2 {:(2x2 +-2x β 4) dx =1%β+%2'β4x1_:=(ββ+4+8)β(β1s+9+12)= #33217; A2: Forthesketchgivema:β2andb=1:f(x)βg(x)=(βx2-2x)β(x2β4)=β2x'2β2x+4 :_ 3 1 => A2=-β r_2(22&iβββ+214β4)dx-=β12~13β-+)12β4x1w_2=β(%+1β4)+(β53~'Γ© +4+8) =βΒ§β1+4β%+4+8=9; Therefore, AREA '2 A1 +A2~_-w 131 +92 % 68. Limits of integration 7β 2x2=x2+4 :> 3x2- 3:0 => 3(xβl)(x+l)=0 : a=β1andb=1; f(x) ~ g(x) m (7 β 2x2) β- (x2 +4) m 3 β- 3x2 =>A= [1(32.3x2)dx=3 1-33β3114 =3[(1-β-)- (-1+ 91:64): 75. Limitsofintegration: 4x=y2β4and4x=16+y =>y2β4ml6+y=>y2βyβ2020=> (yβ5)(y+4)(:_10:>c=β4andd:5; f(y)β g(y)= )-(L-)= "42+ β0 II :4Β»: all-4 and A I H In + NIH + ._. E, I Inn A l3 + I; I oo 3 84. Limits ofintegration: x m 3 β y2 and x z β y; 2 => 3-3122β? a} 379β320 2> %(yβ2)(y+2)m0 => c=β2andd=2;f(y)βg(y)=(3-Y2)β (1:)2 =3(1ββ)=>A= 3r:(1ββ)dy= 3fyβ _2 =3[(2β-2)_(_2+12)]23(4_12)=12_4β2=8 ...
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05.06 - 5.6 SUBSTITUTION AND AREA BETWEEN CURVES 4(a...

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