Mth 6 Clculus II
Contents 7 Additionl topics in Integrtion 7. Integrtion by prts..................................... 7.4 Numericl Integrtion.................................... 7 7.5 Improper Integrl...................................... 9 8 Clculus of Severl Vribles 4 8. Functions of severl vribles................................ 4 8. Prtil Derivtives...................................... 5 8.3 Mxim nd minim of functions of severl vribles................... 7 8.4 Lest Squres........................................ 9 8.5 Constrined Optimiztion nd Lgrnge multipliers.................... 8.6 Constrined Optimiztion nd Lgrnge Multipliers(continued).............. 8.7 Totl Differentils...................................... 3 8.8 Multiple Integrls...................................... 4 8.9 Applictions......................................... 6
Chpter 7 Additionl topics in Integrtion 7. Integrtion by prts Derivtion: u = u(x). du = u dx, ex. d sin x = cos xdx Especilly, d[uv] = udv + vdu = uv dx + vu dx = [uv + vu ]dx = [vu] dx (product rule). Ex. d[cos x sin x] = cos xd sin x + sin xd cos x = cos xdx sin xdx = [cos x sin x]dx = [sin x cos x] dx. duv = udv + vdu, uv = udv + vdu + C udv = uv vdu + C, or uv dx = uv vu dx + C Note: We should choose u,v such tht u v is simpler thn uv. Exmples:. xe x dx. Choose u nd dv, using try-nd-test Cse : u = e x, dv = xdx
. du = e x dx, v = Cse : e x }{{} =u =dv {}}{ xdx = e x }{{} u dv = v {}}{ x xdx = x du x {}}{ e x dx }{{} v u = x, dv = e x dx du = dx, v = e x dx = e x xe x dx = xe x e x dx = xe x e x + C x 4 ln xdx. u = ln x, dv = x 4 dx 3. du = x dx, v = x5 5 x 4 ln xdx = ln(x) x5 5 (x + )(x 5) 5 dx x x5 5 dx 4. u = (x + ), dv = (x 5) 5 dx (x 5)6 du = dx, v = 6 (x + )(x 5) 5 dx = (x + ) x x + dx. (x 5)6 6 (x 5) 6 dx 6 u = x, dv = x + x du = dx, v = + dx = (x + )3/ 3 x x + = x 3 (x + )3/ 3 (x + )3/ dx = 3 x(x + )3/ 4 5 (x + )5/ + C Rewrite it s x(x + ) / dx, then refer to exmple 4. 3
5. ln xdx u = ln x, dv = dx 6. du = x dx, v = x ln xdx = ln(x) x x xdx. (ln x) dx (repeted use by prt) u = (ln x), dv = dx du = ln(x) x, v = x (ln x) dx = (ln x) x u = ln(x), dv = dx ln(x) x xdx = (ln x) x ln(x)dx. du = x dx, v = x ln(x)dx = ln(x)x xdx = ln(x)x x x (ln x) dx = (ln x) x ln(x)x + x. Some tricks. xf(x)dx, usully we choose u = x, f(x)dx, u = x, dv = f(x)dx du = dx, v = f(x)dx = F (x) xf(x)dx = xf (x) F (x)dx Exmples () x sin xdx: F (x) = sin xdx = cos x, F (x) = cos xdx = sin x (b) x cos xdx: (c) xe x dx: F (x) = e x dx = e x, F (x)dx = e x dx = e x 4
(d) (e) x(x+) n dx: F (x) = x ln xdx (x+) n dx = (x + )n+, n + (x + ) n+ dx = n + (x + )n+ (n + )(n + ). () u = lnx, du = x, = x ln x dv = xdx v = xdx = x x x x dx = ln x x f(x)dx: using integrtion-by-prt two times x cos xdx st time: x x x dx = ln x 4 + C u = x, dv = cos xdx du = xdx, v = sin x = x sin x x sin xdx nd time: Clculte x sin xdx u = x, dv = sin xdx du = dx, v = cos x = x cos x ( cos x)dx = x cos x + sin x (b) (c) x cos xdx = x sin x + x cos x sin x + C x cos xdx x e x dx st time: u = x, dv = e x dx du = xdx, v = e x = x e x xe x dx 5
nd time: Clculte xe x dx u = x, dv = e x dx du = dx, v = e x = xe x e x dx = xe x e x (d) (e) x e x dx = x e x xe x + e x x (x + 3) dx x ln xdx u = ln x, dv = x dx du = x dx, v = x3 3 x ln xdx = x3 3 ln x x 3 x 3 dx = x3 x 3 ln x 3 dx (f) (ln x) dx x 3 Need to use times of integrtion by prt. st time: u = (ln x), dv = x dx du = ln x x, v = x = (ln x) x + ln xx dx nd time: clculte ln xx dx u = ln x, dv = x dx du = x dx, v = x = ln xx + x dx = ln xx x 6
Exmple 7: A politicin cn rise cmpign funds t the rte of 5te.t thousnd dollrs per week during the first t weeks of cmpign. Find the verge mount rised during the first 5 weeks. 5 verge = 5 5te.t dt u = 5t, dv = e.t dt du = 5dt, v = e.t. 5te.t dt = 5t e.t 5 5. 5 e.t. dt = 5t e.t 5 5 e.t + 5. 7.4 Numericl Integrtion Integrtion Techniques: from integrtion rules; by substitution; by prts. But we my not be ble to find the closed form of mny integrls. For exmple Approximtion: (Riemnn Sum) e x dx =?, sin(x) dx =? x Consider b The Rectngle Rule: f(x)dx. Given n integer n >, divide [, b] into n subintervls. Then x = b n, x =, x = x + x,..., x n+ = x n + x = b. R n = xf(x ) + xf(x ) + + xf(x n ) b f(x)dx def = lim n R n The following figures show how Reimnn Sum pproximtes integrl when n =,, 3, 6,,. 7
4 4 4 3 3 4 5 6 7 3 3 4 5 6 7 3 3 4 5 6 7 4 4 4 3 3 4 5 6 7 3 3 4 5 6 7 3 3 4 5 6 7 It cn be proved (in Numericl nlysis) tht The Trpezoidl Rule: Simpson s Rule: (Just mention) b f(x)dx R n M (b ) n [ T n = x f(x ) + f(x ) + + f(x n ) + ] f(x n+) b f(x)dx T n M (b ) 3 n S n = x 3 [f(x ) + 4f(x ) + f(x 3 ) + 4f(x 4 ) + + 4f(x n ) + f(x n )] b f(x)dx S n M 3(b ) 5 n 4 These rules cn be esily progrmmed on computer (including your grphic clcultor). Exmple : Use the rectngulr nd the trpezoidl rules to pproximte Soln: x = 3 4 =. 3 dx with n = 4. x 8
Rectngulr Rule x f(x) = /x 3/ /3 / 5/ /5 3 /3 sum=.5667 s =.567 =.83 Trpezoidl Rule x f(x) = /x / 3/ /3 / 5/ /5 3 /3 /6 sum=.33 s =.33 =.67 Error for rectngulr pproximtion: Error for Trpezoidl pproximtion: 3 3 3 x dx = ln(x) 3 = ln 3 x dx S 4 =.84 x dx T 4 =.8 Trpezoidl pproximtion is much more ccurte thn rectngulr pproximtion. 7.5 Improper Integrl It is known tht b x-xis (if f(x), x [, b]). f(x)dx mens the re of the region bounded by y = f(x), x =, x = b nd the How cn we find the re of the region bounded by y = e x, x =, nd x-xis? A = This is not n ordinry integrl. How to clculte it? Ide: Approximtion: let b is constnt nd lrge. b e x dx = ex e x dx b = e b. 9
Tking limit on both sides, It is resonble to regrd Definition: The improper integrl lim b b e x dx = lim b ( e b ) = e x dx = lim b b f(x)dx is defined s e x dx. Similrly b f(x)dx = lim b f(x)dx = b b lim f(x)dx. f(x)dx, f(x)dx = = lim f(x)dx + f(x)dx f(x)dx + lim b b f(x)dx An improper integrl is sid to be convergent if the limit (or limits) exists nd to be divergent if the limit (or limits) does not exist. L Hôpitl s rule lim x c f(x) = lim x c g(x) =, or ± lim x c f (x) g (x) exists, nd g (x) for ll x c. f(x) then lim x c g(x) = lim f (x) x c g (x) Exmples. lim x x e x = lim x e x = x. lim x e x = lim x x e x = lim x e x = sin x 3. lim x x = lim cos x = cos() = x
4. lim x e x ln x = lim x e x /x = lim x xex =. Limits of some specil functions e = lim x ex e = lim x ex ln( ) = lim ln x x ln() = lim ln x x lim x /xn n > lim x /xn n > lim x xe x lim x xn e x, n >, Exmples.. x + dx = ln( x + ) x (x + ) dx. = (divergent). Using method of substitution. Let u = x +, then du = xdx, xdx = du x (x + ) dx = du u = ln( u ) 5 = + 3. 4. 5. x dx = x = [ ] [ ] = + =. x dx = x =. xe x dxdx using integrtion by prts u = x, dv = e x dx du = dx, v = e x xe x dx = xe x + e x dx = xe x e x xe x dxdx = xe x e x
lim x xe x = lim x x e x = x, xe x e x =, x, xe x e x = Soln= ( ) = Improper Integrl: f(x)dx Let c be ny rel number nd suppose both the improper integrls re convergent. Then the improper integrl c f(x)dx nd c f(x)dx f(x)dx = c f(x)dx + c f(x)dx Exmples. xe x dx xe x dx = using method of substitution xe x dxdx + xe x dxdx u = x, du = xdx xe x dx = e u du = eu = e x x, e x = x, e x = xe x dxdx = e x = x, e x = xe x dxdx = e x = ( ) = xe x dx = + =
Review of Chpter 7 Integrtion by prts: uv dx = udv = uv vdu = uv vu dx Improper integrls: Numericl integrtion: Rectngulr rule. trpezoidl rule. Simpson s rule*. Exmples:. Integrtion by prts. f(x)dx, ln(t) t dt. b inf f(x)dx, f(x)dx. u = ln(t), dv = t dt = t / dt. Integrtion by prts. du = dt, v = t/ t ln(t) dt = ln(t) t t t t/ dt (x + 3)(x ) 4 dx u = (x + ), dv = (x ) 4 dx 3. Substitution. (x )5 du = dx, v = 5 (x + 3)(x ) 4 dx = e x ( + e x ) 3. (x + )(x )5 5 (x ) 4 dx u = + e x, du = e x dx. e x ( + e x ) 3 = u du = u3 = 3
Chpter 8 Clculus of Severl Vribles 8. Functions of severl vribles Definition(function):(y = f(x)) A function is rule such tht to ech vlue x in the domin, there corresponds one nd only one number y. Wht to know for y = f(x): find the domin: set of ll x for which f(x) is defined. nturl domin: lrgest set of ll x for which f(x) is defined. Rnge: the set of ll resulting vlues of the function sketch the grph (rnge, domin, specil points, behvior s x. differentite f(x). integrte f(x). Exmple: f(x) = x domin: x or {x x } = [, + ). rnge: {y y } = [, + ). y = x x y = x.44 4 3 4 5 4
Functions of Severl Vribles : Motivtion: temperture depends on plce (ltitude, longitude, nd elevtion) nd time T = T (t, θ, φ, h) Def: A function f of two vribles is rule such tht to ech ordered pir (x, y) in the domin of f, there corresponds one nd only one number f(x, y). or z = f(x, y). x Exmple : f(x, y) = y Domin: D = {(x, y) x, y > } Rnge: z Exmple : f(x, y) = x, f(, e), f(e, )? ln(y) (x, y) z = f(x, y) Exmple 3: f(x, y) = x y, f(x, y) = ln(x + y ). Exmple 4: z = 8 x y Exmple 5: z = y x 8. Prtil Derivtives Def (Using prtil derivtives): the rte of chnge of function with respect to one vrible while holding ll other vribles constnt. Consider differentiting f(x) Wht bout two vribles (x, y) df dx = lim f(x + h) f(x) h h f(x) = cx 3, c is constnt df dx = d dx (cx3 ) = 3cx f(x, y) = x 4 + y d dx (x4 + y ) = 4x 3 y held constnt d dy (x4 + y ) = y x held constnt 5
Nottions: f x = f (x, y) = f(x, y) = d x x f y = f y (x, y) = f(x, y) = d y f x f (x, y) = lim x h dx dy f(x, y) f(x, y) f(x + h, y) f(x, y) (x, y) = lim h h f(x, y + h) f(x, y) h y held constnt x held constnt Exmples: f(x, y) = x 3 + 3x y y 3 x + y f x = 3x + 6xy + f y = + 6x y 6y + f(x, y) = ln( x + y ) f x = f y = x + y x + y (x) x + y x + y (y) w = (u v) 3 f(x, y) = e x +y, find f x (, ), f y (, ) w = 3(u v) u w v = 3(u v) ( ) f x = e x +y (x), f x (, ) = e ( ) = f y = e x +y (y), f y (, ) = e ( ) = e Higher Order Derivtives: f = f(x, y), f xy = ( ) f. Similr rules re pplied to f xx, f xy, f yx y x Exmple: Second-order derivtives of f(x, y) = 5x 3 x y 3 + 3y 4 6
Soln: f x = 5x 4xy 3, f y = 6x y + y 3 f xx = x (5x 4xy 3 ) = 3x 4y 3, f yx = f xy = y (5x 4xy 3 ) = xy f yy = y ( 6x y + y 3 ) = x y + 36y 8.3 Mxim nd minim of functions of severl vribles Introduction: minimum points, mximum points, sddle points. How to define or chrcterize them? Criticl points (Def): (, b) is criticl point of f(x, y) if f x (, b) =, nd f y (, b) = Reltive mximum nd minimum vlues cn occur only t criticl points. reltive mximum, minimum nd sddle points re criticl points. Exmple : find criticl points of f(x, y) = 3x + y + xy + 8x + 4y. Soln: { fx (x, y) = 6x + y + 8 = ; f y (x, y) = 4y + x + 4 = x = 6 5, y = 5 Second Derivtive test for functions f(x, y) The D-test: (, b) is criticl point of f(x, y). Let D = f xx (, b) f yy (, b) [f xy (, b)], (i) if D > nd f xx (, b) >, f(x, y) hs reltive (locl) minimum t (, b); (ii) if D > nd f xx (, b) <, f(x, y) hs reltive (locl) mximum t (, b); (iii) if D <, f(x, y) hs sddle point t (, b). (iv) if D =, inconclusive. (, b) cn be either reltive mximum, reltive minimum or sddle point. 7
Def: (Sddle Point): sddle point is sttionry point (criticl point) but not locl extremum. A criticl point is either locl minimum, locl mximum or sddle point. Exmple : f(x, y) = 3x + y + xy + 8x + 4y, f xx = 6, f xy =, f yy = 4. D = f xx f yy f xy = 4 4 = >, f xx >, reltive minimum Exmple 3: find the reltive extreme vlues of f(x, y) = e 5(x +y ). f x (x, y) = xe 5(x +y ) f y (x, y) = ye 5(x +y ) f xx (x, y) = e 5(x +y )+x e 5(x +y ) f xy (x, y) = xye 5(x +y ) f yy (x, y) = e 5(x +y ) + y e 5(x +y ) Criticl Points: { xe 5(x +y ) = ye 5(x +y6) = x =, y =. f xx (, ) = > f xy (, ) = f yy (, ) = D = = > (, b) = (, ), minimum points f(, ) =. Exmple 4 f(x, y) = x 3 y 3 3x + 6y. 8
8.4 Lest Squres We wnt to find stright line to fit these dt. 5 E G 5 E G F H F H 5 5 Exmple We try to find line y = x + b to best fit the following 4 given points. Let We find, b by x y x+b error=x+b-y 6 +b +b-6 6 6+b 6+b- 5 +b +b-5 +b +b- S(, b) = ( + b 6) + (6 + b ) + ( + b 5) + ( + b ) Generl Cse: fit stright line to dt min S(, b),b x y xy x x y x y x x y x y x.... x n x y n y x n y n xy x n x the lest squre line is y = x + b = n xy ( x)( y) n x ( x) b = n ( y x) Exmple : n = 3. 9
x y xy x 4 4 3 5 75 9 x = 6 y = 47 xy = 9 x = 4 fitting line: y = 7.5x +.67 = n xy ( x)( y) n x ( x) 3 9 6 47 = 3 4 6 = 7.5 b = n ( y x) = (47 7.5 6) =.667 3 C A B 3 Exmple Fit stright line to () x y. 3.5 35 38.5 4 (b) x y 5 8 8 3
8.5 Constrined Optimiztion nd Lgrnge multipliers Review: unconstrined Optimiztion: f x = f y = Constrined Optimiztion z = f(x, y) = x + y f f = x, x y = y (x, y) = (, ) (Criticl point nd minimum point) Find the minimum of the intersection curve between z = x + y nd y = plne, which is equivlent to minimize x + y, subject to y =. missing grph here Method : Method : Let (x, y) = (, ) f(x, y) = x + y, f(x, ) = x + 4 minimize(x + 4) x = is the minimum point, nd the corresponding minimum function vlue is 4 F (x, y, λ) = x + y + λ(y ) F x = x, x = F = y + λ, y + λ = y F λ = y, y = x =, y =, λ = 4 (x, y) = (, ) is the minimum point with minimum vlue f(, ) = 4. Method of Lgrnge multipliers minimize f(x, y) subject to g(x, y) = (constrint) (i) define F (x, y, λ) = f(x, y) + λg(x, y).
(ii) Find criticl points (x, y, λ) of F : F x =, F y =, F λ =. (iii) The solutions found in step re cndidtes for the extrem of f. Note: Of the method of Lgrnge multipliers, there is no criterion to determine whether criticl point of function of two or more vribles leds to reltive mximum or reltive minimum. Exmple f(x, y, z) = xy + 6yz + 8xz, constrint: xyz =,. 8.6 Constrined Optimiztion nd Lgrnge Multipliers(continued) Exmple : A continer compny wnts to design nd luminum cn requiring the lest mount of luminum, but tht contins exctly 6π cubic inches. Find the rdius nd the height of the cn. A = πr + πr h V = πr h min r,h πr + πr h subject to πr h = 6π Soln: Using method of Lgrnge multiplier F (r, h, λ) = πr + πrh + λ(πr h 3π) F r = 4πr + πh + πrhλ = () F h = πr + λπr = () F λ = πr h 6π (3) From () λr = λr in (), 4πr πh =, r = h r h in (3), πr 3 6π =, r = r =, h = 4 The minimum mount of luminum needed is π() + π()(4) = 4π, when r =, h = 4.
Exmple :Minimize or mximize f(x, y) = xy, subject to x + y = 8. F (x, y, λ) = xy + λ(x + y 8). F x = y + xλ = ; () F y = x + yλ = ; () F λ = x + y 8 =. (3) y(y + xλ) = (3) x(x + yλ) = (4) (3) (4) y x =. Criticl points: { y x = x + y 8 = y = ±3 x = ±3 ( 3, 3) ( 3, 3) (3, 3) (3, 3) f( 3, 3) = 8; f( 3, 3) = 8; f(3, 3) = 8; f(3, 3) = 8; minimum: 8, mximum: 8. More Exmples: Mximize nd minimize f(x, y) = x + 3y, subject to x + 5y = 8. 8.7 Totl Differentils One-vrible function f(x), chnge in x: dx. x x + dx f(x) f(x + dx) f = f(x + dx) f(x) f (x)dx. Two-vrible function f(x), chnge in x: dx, chnge in y: dy. f = f(x + dx, y + dy) f(x, y) = f(x + dx, y + dy) f(x, y + dy) + f(x, y + dy) f(x, y) f x (x, y + dy)dx + f y (x, y)dy f x dx + f y dy. df f x dx + f y dy f Approximte chnge of f Exmple: f(x, y) = e 3x y df = e 3x y 3dx + e 3x y ( )dy 3
Exmple : f(x, y) = ln( + x + y ). df =? Exmple : f(x, y) = x y, find f when (x, y) chnges from ( 3, ) to ( 3.,.98) x + y dx = 3. ( 3) =., dy =.98 ( ) =. f x = (x y) x(x + y) (x + y) x(x y) (x + y) = y (x + y) f y = (x y) y(x + y) (x + y) y(x y) (x + y) = x (x + y) f df = f x dx + f y dy x= 3,y=,dx=.,dy=. =.8/5 +./5 =./5 =.8 Applied Exmple: (Approximting Chnges) Find the pproximte chnge in the volume of cylinder when the rdius is incresed from 5 to 6 nd the height is decresed from 4 to. V = πr h dv = V V dx + r h dh r = 5, h = 4 dr =, dh = = (πrh)dr + (πr )dh r = 5, h = 4 dr =, dh = = 4π 5π = π 8.8 Multiple Integrls Definition:(Double integrl) The double integrl of continuous function f(x, y) on rectngulr region R is f(x, y)dxdy = lim f(xi, y j ) x y x, y (The sum is over ll rectngles in R) R If f(x, y) is non-negtive on R, then the double integrl gives the volume under f over R. Definition: (Iterted integrls) R is the rectngle defined by x b, c y d nd re clled iterted integrls. d c [ b ] f(x, y)dx dy, b [ d c ] f(x, y)dy dx 4
Exmple : Exmple : Rule : d c [ b 8xydydx = 4 ] f(x, y)dx dy = [ (x + y )dxdy b [ d Rule : Let R = {(x, y) x b, c y d} d b f(x, y)dxdy = f(x, y)dxdy = R Exmple 4: R : x, y, f(x, y) = c c xy + y. Exmple 5: R = {(x, y) x, y } 6x ydxdy. Double integrl over non-rectngulr region R ] 8xydy dx ] f(x, y)dy dx b d c f(x, y)dydx Exmple 6 R = {(x, y) x, y e x }, f(x, y) = xy Slon: e x e x xydydx xydy = /xy ex = /x(e x ) /x = /xe x /xe x dx, Using integrtion by prts u = /x, dv = e x dx du = /dx, v = e x dx = /e x. F (x) = /xe x dx = (/x)(/e x ) /4e x dx = /4xe x /8e x F () F () = /4e /8. Note: since the region of y depends on x, the order of the integrl is NOT exchngeble, tht is, we cn not clculte e x ( ) xydx dy. 5
Exmple : f(x, y) = y + x ; R is the region bounded by y = x, y =, nd x = 4. Region R 4 3 R B A 3 4 5 6 Soln: 4 x y + x dydx x 4 y y dx = + x + x x = x + x x dx, Using method substitution + x u = ( + x ), du = xdx 4 x + x dx = / u du = / ln u = / ln( + x ) 4 = / ln 5 8.9 Applictions Averge: Exmple 4: f(x, y) = + 6x y Averge vlue of f over R = f(x, y)dxdy re of R R R = {(x, y) x, y 3} 6
Exmple Suppose the popultion of certin city is f(x, y) =, e.x.y /mile Let (, ) gives the loction of the city hll, wht is the popultion inside the rectngulr re described by R = {(x, y) x ; y 5} Soln: The popultion is 5 5, e.x.y dydx, e.x.y dy =, e.x.y 5 =, [e.x.5 e.x ] (, )[e.x.5 e.x ]dx = 5, [e.x.5 e.x ] = 5, [e.5 e e.5 + ] = 79.5 Review of Chpter 8 functions of severl vribles: definition, domin, nturl domin, rnge, three dimensionl coordinte system. prtil derivtive (f x, f y, f xx, f yy,...) Optimiztion problem: Criticl points (reltive minimum points, reltive mximum points, sddle points) D-test Constrined Optimiztion problem: Lgrnge multiplier method. Lest squres: stright line fitting totl differentils double integrtion: iterted integrls, verge over domin. Exmple : Find ll reltive extreme vlues: nd tell if they re minimum or mximum. f(x, y) = xy x 5y + x y + 3 Exmple : Stright line fitting x y 7 4 4 5-7
Exmple 3: = n xy ( x)( y) n x ( x), b = n ( y x) mximize f(x, y) = 4xy x y subject to x + y = 6 Exmple 4:A compny s profit is p = 3x /3 y /3, where x nd y re respectively, the mounts spent on production nd dvertising. The compny hs totl 6, dollrs to spend. find the mounts for production nd dvertising tht mximizing the profit. 8