Calculus II Solutions review final problems

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1 Calculus II Solutions rviw final problms MTH 5 Dcmbr 9, 007. B abl to utiliz all tchniqus of intgration to solv both dfinit and indfinit intgrals. Hr ar som intgrals for practic. Good luck stuing!!! (a) u = x x x Lt du = 9x 9 du = x x x = 9 u du = 9 x + C (b) u = ln(x) dv = x x ln(x) Lt du = x v = x x ln(x) = x ln(x) x x = x x ln(x) ( ) = ( x x ln(x) 9 = 6 ln( ) 6 9 ln() + 9 (c) u = sin(x x sin(x )cos(x ) Lt ) du = x cos(x ) x sin(x )cos(x ) = udu = 6 sin (x ) + C (d) sin ( θ) ( dθ = ( cos 4 θ) [ )dθ = θ 4 sin ( 4 θ)] + C. () 4 4 sin( t) t Lt sin( t) (f) p sc (π p ) tan(π p ) dp (g) u = t du = t du = t t = sin(u)du = cos(u) = cos( t ) 4 = [cos( ) ] u = tan(π p ) Lt du = p sc (π p )dp du = p sc (π p )dp du u = ln u = ln tan(π p ) + C p sc (π p ) tan(π p ) dp = x 7 x = sin(θ) x Lt = cos(θ) 9 x = cos(θ) 9 x = 9 cos (θ) x 7 = sin(θ) 7 9 x cos(θ) cos(θ)dθ = ( sin(θ) 7)dθ = cos(θ) 7θ = (h) x 6 x x 6 x Lt = 9 x 7 arcsin( x ) + C x = 4 sc(θ) x = 6 sc (θ) = 4 sc(θ)tan(θ)dθ x 6 = 6 tan (θ) = 4tan(θ) 4sc(θ)tan(θ)dθ 4sc(θ) = tan (θ)dθ = (sc (θ) )dθ = [tan(θ) θ] = [ x 6 4 sc ( x 4 )] + C

2 x +0x+ x(x +) = A x + Bx+C x + (i) x + 0x + = A(x + ) + (Bx + C)x x +0x+ x(x +) hr if x = 0; A = cofficints of highst dgr trms x ; A + B = B = 0 x; C + B = 0 C = 0 x +0x+ x(x +) = x + 0 x + = ln(x) + 0 arctan(x) + C (j) x u = 9 + x u 9 = x 9 + x Lt du = x du = x [ ] x 9 + x x = (u 9)u du = 5 u 5 6u = 5 (9+x ) 5 (9+x ) +C (k) x+ (x+)(x ) hr x+ (x+)(x ) = [ 4 x+ + x x+ (x+)(x+)(x ) = A x+ + B x+ + C x x + = A(x + )(x ) + B(x + )(x ) + C(x + )(x + ) if x = ; C = 5 6 if x = ; B = if x = ; A = 4 x ] = 4 ln(x+)+ ln(x+)+ 5 6 ln(x )+C (l) sin (θ)cos (θ)dθ = sin (θ)( sin (θ))cos(θ)dθ Lt [ ] u ( u )du = sin (θ) sin5 (θ) 5 + C (m) x = sin(θ) x x Lt = 4 sin (θ) 4 x = cos(θ)dθ 4 x = 4 cos (θ) cos(θ)dθ 4 sin (θ) cos(θ) = 4 csc (θ)dθ = 4 cot(θ) = x 4 + C 4 x (b) π 0 cos(θ) sin(θ) dθ = lim b 0 π b cos(θ) sin(θ) dθ u = sin(θ) du = cos(θ)dθ. Evaluat th following intgrals: (a) u = ln(x) hr x ln 9 x du = x u 9 du x ln 9 x = lim b b ln 8 x b = lim b 8 ln 8 b + 8 = 8 Convrgs π cos(θ) 0 dθ = lim b 0 π [ sin(θ) = lim b 0 sin(θ) b Convrgs. Considr th intgral (x ) lft hand ndpoint of th intrval at x =. lim b + b (x ) = lim b + u = sin(θ) du = cos(θ)dθ u du = u ] sin π b =. This intgral is impropr sinc it is undfind at th x = lim b +[ + b b ] divrgs

3 4. Writ th quation for th volum obtaind by rotating th rgion shadd blow about th lin y = Volum brokn into two parts using washrs: V = [ ((x 0 π x + x) ( ) ) (x ( )) ] + π [ (x ( )) ( (x x + x) ( ) ) ] 5. Sktch and find th ara btwn th curvs y = 4x x and y = x Intrsction points: x = 4x x 0 = x x 0 = x( x) x = 0, x = ytop y bottom = 0 (4x x x) = 0 (x x ) = x x 0 = 4 units

4 6. Find th volum of th solid objct gnratd whn y = x + 5, x =, x = 0 is rvolvd about th y-axis. Tak on cylindr, thn cut and unroll. Radius of ach shll: x Circumfrnc of dg: πx. Hight of ach shll: y top y bottom = x + 5. Thicknss:. on small rctangular slab volum: dv = πx( x + 5). Total volum: V = 0 πx( x + 5) = π 0 ( x + 5x) = π [ x + 5 x] 0 = π [ (8) + 5 (4)] = π[ 4+0] = πunits. 7. Suppos that A and B ar constants. Vrify that y(x) = Ax + B + ln(x) is a solution of th diffrntial quation xy + y = x y(x) = Ax + B + ln(x) y = Ax + x y = Ax x Substituting into th DE w gt: x(ax x ) + ( Ax + x ) Ax x Ax + x = x = x 8. Tru or Fals: Th diffrntial quation givn by + xy = has a solution of x y(x) = x 0 t + C x This problm was don in class. 9. Givn a diffrntial quation of th form = kxy, find constant k such that y = is a solution to this DE. From y = x +5 y = x (x +5) = kxy x (x +5) = kx (x +5) k = 0. Solv th IVP for th spcific solution, = y+ Sprat variabls: y+ = t 4 t 4 ; y(0) = ln y + = ln t 4 + C y + = A(t 4) y = A(t 4) Applying th initial condition; y(0) = 4A = A = 5 4 Spcific Solution is: y(t) = 5 4 (t 4) y(t) = 5 4 t x +5 4

5 . Solv th IVP for th spcific solution, x + xy = y; y() = x = y xy = y( x) y = x x = ( x ) lny = lnx x + C y = ln x x+c = ln x x C y(x) = Ax x Applying th initial condition: y() = A = A = = Spcific solution: y(x) = x x = x x. A vat initially contains 00 gallons of sugar watr with tablspoons of sugar pr gallon. Sugar watr containing 6 tablspoons of sugar pr gallon ntrs th tank at th rat of 5 gallons pr minut. Wll-mixd sugar watr drains from th bottom of th vat, kping th total volum constant. Lt S(t) dnot th numbr of tablspoons of sugar in th vat at tim t minuts. Find an xprssion for S(t) in trms of t. = Rat in - Rat out Rat in = (Concntration In)(Flow Rat In) = 6 tbsp. gal 5 gal min = 0tbsp. min Rat out = (Concntration Out)(Flow Rat Out) = S tbsp. 00 gal 5 gal min = S tbsp. 60 min = 0 S 60 with initial condition S(0) = 600 = 60 (S 800) S 800 = 60 S 800 = 60 ln S 800 = t 60 + C S(t) = 800+A 60 t applying initial condition S(0) = 600 = 800+C C = 00 S(t) = t. Suppos th population of minks dvlops according to th quation dp = P +7P 0P whr t is in yars and P is in thousands of minks. What ar th quilibrium solutions? Basd on th slop filds classify th quilibrium solutions as stabl/unstabl. If P(0) = 4, what happns to dp as t? If P(0) = 4, what happns to P as t? Th quilibrium solution occur whr dp = 0 P + 7P 0P = 0 P(P 7P + 0P) = 0 P(P )(P 5) = 0 P = 0, P =, P = 5 If P =, thn dp dp > 0. If P =, thn < 0 This shows th P = 0 is a stabl quilibrium. If P =, thn dp > 0. If P = 6, thn dp < 0 This shows th P = is an unstabl quilibrium. and P = 5 is a stabl quilibrium. If P(0) = 4, thn dp is positiv and hading towards 0. If P(0) = 4, thn P approachs 5 as tim gos on and on. 5

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