# 2414 Exam 1 Review - solutions(4).pdf - Exam 1 Review...

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Exam 1 Review - Solutions - Page 1/41.r=p(4 + 2)2+ (3-1)2+ (8-5)2=62+ 22+ 32=36 + 4 + 9 =49 = 7;Sphere: (x+ 2)2+ (y-1)2+ (z-5)2= 492.Center:(-5+32,2+82,0-42)= (-1,5,-2),Radius:p(-1-3)2+ (5-8)2+ (-2-(-4))2=16 + 9 + 4 =29,Equation: (x+ 1)2+ (y-5)2+ (z+ 2)2= 293.x2+y2+z2-8x+ 6y-2z-10 = 0; (x2-8x) + (y2+ 6y) + (z2-2z) = 10;(x2-8x+ 16-16) + (y2+ 6y+ 9-9) + (z2-2z+ 1-1) = 10; (x2-8x+ 16) + (y2+ 6y+ 9) + (z2-2z+ 1)-26 = 10;(x-4)2+ (y+ 3)2+ (z-1)2= 36.Center: (4,-3,1), Radius: 64.h6 + 2,4-5,-5 + 3i=h8,-1,-2i5.|a|=p82+ 42+ (-1)2=64 + 16 + 1 =81 = 9; 7a|a|=79a=h569,289,-79i6.|-→PQ|=p(6 + 2)2+ (4-5)2+ (-5 + 3)2=p82+ (-1)2+ (-2)2=64 + 1 + 4 =69;b|-→P Q=169b=h669,0,-569i7.4a-3b=h4(8)-3(6),4(4)-3(0),4(-1)-3(-5)i=h32-18,16-0,-4 + 15i=h14,16,11i8.6b+ 2-→PQ=h6(6) + 2(8),6(0) + 2(-1),6(-5) + 2(-2)i=h36 + 16,0-2,-30-4i=h52,-2,-34i;6b+ 2-→PQ= 52i-2j-34kNote that on 8 - 12, the position may be different from what you have done. As long as the magnitude and directionare the same, then it is correct.9.10.11.12.13.14.Z12x2+ 3x+ 1dx=Z1(2x+ 1)(x+ 1)dx;1(2x+ 1)(x+ 1)=A2x+ 1+Bx+ 1; 1 =A(x+ 1) +B(2x+ 1);Ifx=-1,thenB=-1.Ifx=-12,thenA= 2;Z1(2x+ 1)(x+ 1)dx=Z22x+ 1-1x+ 1dx= ln|2x+ 1| -ln|x+ 1|+C15.Integration by parts,u= lnx, dv=x5dx,sodu=1xdx, v=16x6;Z21x5lnx dx=16x6lnx21-Z2116x5dx=16x6lnx-136x621=(646ln 2-6436)-(0-136)=646ln 2-6336
Exam 1 Review - Solutions - Page 2/416.Zπ/20sin3θcos2θ dθ=Zπ/20sin2θcos2θsinθ dθ=Zπ/20(1-cos2θ) cos2θsinθ dθ;u= cosθ, du=-sinθdθ,Ifθ= 0,thenu= 1.Ifθ=π2,thenu= 0.Zπ/20sin2θcos2θsinθ dθ=-Z01(1-u2)u2du=Z10u2-u4du=13u3-15u510=13-15=21517.Integration by parts,u= 3x-1, dv= sin(2x)dx,sodu= 3dx, v=-12cos(2x);Zπ/60(3x-1) sin(2x)dx=-12(3x-1) cos(2x)π/60+32Zπ/60cos(2x)dx=-12(3x-1) cos(2x) +34sin(2x)π/60=-π8+14+338-12=-π8+338-1418.u= arctanx,dudx=11+x2.Ifx= 0,thenu= 0.Ifx= 1,thenu=π4.