10 Διάλεξη - eclass.tuc.gr · 18/10/2016 ©2015 Γιάννης...

10η Διάλεξη Διαφορικά – Αόριστα Ολοκληρώµατα

Κανόνες Ολοκλήρωσης

18 Οκτωβρίου 2016

Γιάννης Σαριδάκης Σχολή Μ.Π.Δ., Πολυτεχνείο Κρήτης

ΑΠΕΙΡΟΣΤΙΚΟΣ ΛΟΓΙΣΜΟΣ, ΤΟΜΟΣ Ι - Finney R.L. / Weir M.D. / Giordano F.R.Πανεπιστημιακές Εκδόσεις Κρήτης

752 Chapter 14: PartiaL Derivatives

w

FIGURE 14.10 Thisgrapbshowsthe seasonal variation of the temperatuIe below ground as a fraction of surface tmnpenrtmc (Exampl' 5).

(a) z - sin x + 2 sin]

with part of its boundary removed or a solid cube with a missing face, edge, or comer point is neither open nor closed.

Functions of more than three independent variables are also important. For exam-ple, the temperature on a surface in space may depend not only on the location of the point p(x, y, z) on the swface but also on the time 1 when it is visited, 80 we would write T = /(x, y, z, t).

computer Graphing Three-dimensional graphing programs for computers and calculators make it possible to graph functions of two variables with only a few keystrokes. We can often get information more quickly from a graph than from a formula.

EXAMPLE 5 The temperature w beneath the Earth's surface is a function of the depth x beneath the swface and the time t of the year. If we measure x in feet and t as the number of days elapsed from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function

w = 008(1.7 X 10-21 -

(The temperature at 0 ft is scaled to vary from + 1 to -1, so that the variation atx feet can be interpreted as a fraction of the variation at the surface.)

Figure 14.10 shows a graph of the function. At a depth of 15 ft, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 25 ft, there is almost no variation during the year.

The graph also shows that the temperature 15 ft below the surface is about half a year out of phase with the surface temperature. When the temperature is lowt:st on the surface (late January, say), it is at its highest 15 ft below. Fifteen feet below the ground, the seasons are reversed. _

Figure 14.11 shows computer-generated graphs of a number of functions of two vari-ables together with their level curves.

y

x

(b) l - (4x2 + ]2.}o!-r-r

FIGURE 14.11 Computer-generated graphs and level curves of typical functions of two variables.

18/10/2016 ©2015 Γιάννης Σαριδάκης/ΕΕΜΗΥ/Πολυτεχνείο Κρήτης 2

Κανόνας L’Hopital

Απροσδιόριστες µορφές και ο κανόνας του l’Hôpital (Bernoulli)

Εάν f (a)= g(a)=0 και υπάρχουν οι ′f (a) και ′g (a)≠0 τότε

limx→a

f (x)g(x)

= ′f (a)′g (a)

Θεώρηµα 1

( ) ( )( ) ( ) 0 ( ) ( ) ( )lim lim lim lim ( ) ( )( ) ( ) 0 ( ) ( ) ( )x a x a x a x a

f x f af x f x f x f a f ax a

g x g ag x g x g x g a g ax a

→ → → →

−′− − −= = = =− ′− −

−18/10/2016 ©2015 Γιάννης Σαριδάκης/ΕΕΜΗΥ/Πολυτεχνείο Κρήτης 3

1

2

1

2

( )( )

( )( )

( ) ( )lim lim( ) ( )x a x a

f xmf x x

g xg x mx

mf x f ag x m g a→ →

Δ= =

Δ⇓

′= =

′


Εάν ή και και διαφορίσιµες

στο διάστηµα µε για όλα τα τότε

εφόσον το όριο στο δεξιό µέλος υπάρχει.

lim ( ) lim ( ) 0

( , ) ( ) 0 ( , ) \{ }( ) ( )lim lim( ) ( )

x a x a

x a x a

f x g x f g

c d g x x c d af x f xg x g x

→ →

→ →

= = ±∞

′ ≠ ∈′=′

Θεώρηµα 3

Εάν και και διαφορίσιµες στο διάστηµα

µε για όλα τα τότε

εφόσον το όριο στο δεξιό µέλος υπάρχει.

( ) ( ) 0( , ) ( ) 0 ( , ) \{ }

( ) ( )lim lim( ) ( )x a x a

f a g a f gc d g x x c d a

f x f xg x g x→ →

= =′ ≠ ∈

′=′

Θεώρηµα 2



Θεώρημα Μέσης Τιμής του Cauchy

Υποθέστε ότι οι συναρτήσεις f(x) και g(x) είναι συνεχείς στο [a,b] και διαφορίσιμες στο (a,b) με g΄(x)≠0 για κάθε x στο (a,b). Τότε υπάρχει c στο (a,b) τέτοιο ώστε:

′f (c)′g (c)

= f (b)− f (a)g(b)− g(a)

ΑπόδειξηΠαρατηρήστε ότι από το Θεώρημα Μέσης Τιμής. Στη συνέχεια θεωρήστε τη συνάρτηση

και παρατηρήστε ότι F(a)=F(b)=0. Συνεπώς υπάρχει c στο (a,b) τέτοιο ώστε:

′g (x) ≠ 0⇒ g(a) ≠ g(b)

F(x) = f (x)− f (a)− f (b)− f (a)

g(b)− g(a)g(x)− g(a)⎡⎣ ⎤⎦

′F (c) = 0⇒ ′f (c)− f (b)− f (a)

g(b)− g(a)′g (c) = 0 ο.ε .δ .

Η απαίτηση ύπαρξης του ορίου είναι ουσιαστική:

ενώ

sin 1 coslim lim 1 limcos1

sinsinlim lim 1 1

x x x

x x

x x x xx

xx xxx

→∞ →∞ →∞

→∞ →∞

+ += = +

⎛ ⎞+ = + =⎜ ⎟⎝ ⎠

0 0

0 0 0

2

Λανθασµένη εφαρµογή παραβιάζοντας τις προϋποθέσεις:

2

1 cos sin 0lim lim 01 2 1

1 cos sin cos 1lim lim lim1 2 2 2

x x

x x x

x xxx x

x x xxx x

→ →

→ → →

− = = =++

− = = =++


0 0

0 0

Ο κανόνας εφαρµόζεται και µε πλευρικά όρια:

2

ενώ

2

sin coslim lim2

sin coslim lim2

x x

x x

x xxx

x xxx

+ +

− −

→ →

→ →

= =∞

= =−∞

( )1 1 1

1 1

Και άλλες απροσδιόριστες µορφές µπορούν υπολογισθούν µε τον κανόνα:

1 ln 1 lnlim lim lim11 ln 1 ln ln

ln 1 ln 1lim lim1 ln 2 ln 2

x x x

x x

x x x x xxx x x x xx

x x xx x x x

→ → →

→ →

⎛ ⎞⎜ ⎟ =⎜ ⎟⎝ ⎠

− +− = =−− − ⎛ ⎞+⎜ ⎟⎝ ⎠

+= = =− + +


Επιπλοκές:

ενώ2 2

2 2

lim lim lim ...

1 2lim lim lim 11 2

x x x

x x x

x x x x x x

x x x x x x

x x x x

x x x x

e e e e e ee e e e e e

e e e ee e e e

→∞ →∞ →∞

→∞ →∞ →∞

− − −

− − −

−

−

+ − += = =− + −

+ += = =− −

0 0 0 0

0 0 0

Και άλλες απροσδιόριστες µορφές µπορούν υπολογισθούν µε τον κανόνα:

2

ln ln 0

ln 1/lim ln lim lim lim( ) 01/ 1/

lim lim lim 1

x x x x

x x x

xx x x x

x xx x xx x

x e e e

+ + + +

+ + +

→ → → →

→ → →

= = = − =−

= = = =



limx→0+

1+x( )1/x= lim

x→∞1+1

x⎛⎝⎜

⎞⎠⎟

x

= limx→∞

ex ln(1+x−1)

limx→∞

xln(1+ x−1)= limx→∞

ln(1+x−1)1/x

= limx→∞

11+x−1

=1

limx→0+

1+x( )1/x= lim

x→∞1+1

x⎛⎝⎜

⎞⎠⎟

x

= e


Διαφορικά

Διαφορικά


df ≡ df (x) ≡ df (x,Δx) ! ′f (x)Δx

dx ≡ dx(x) ≡ dx(x,Δx) ! (x ′) Δx = Δx df = ′f (x)dx

d 2 f = d(df ) = d ′f (x)dx( ) = ′′f (x)dx2

dn f = f (n) (x)dxn

d(λ f +κ g) = λdf +κ dg

d( f g) = f dg + g df

df (u) = ′f (u)du = ′f (u) ′u (x)dx

Γραμμικοποίηση και Διαφορικά


x 1+ x 1+ x2

Error

0.2 1.095445 1.06 0.03

0.05 1.024695 1.025 0.0003

0.005 1.002497 1.0025 0.000003



Δf = f (a + Δx)− f (a)

≈ L(a + Δx)− f (a) = f (a)+ ′f (a)(a + Δx − a)− f (a)

= ′f (a)Δx = dfx=a


Δf = f (a + Δx)− f (a) ≈

προσεγγίζεται

′f (a)Δx = dfx=a

Σϕάλµα E ! Δf − df

= f (a + Δx)− f (a)− ′f (a)Δx

= f (a + Δx)− f (a)Δx

− ′f (a)⎛⎝⎜

⎞⎠⎟Δx

= ε Δx , ε → 0 καθώς Δx → 0

Δf = ′f (a)Δx + ε Δx = df + ε Δx , ε → 0 καθώς Δx → 0

Το διαφορικό είναι µία εκτίµηση της µεταβολής


Ολοκλήρωση

df (x) = ′f (x)dx ⎯→⎯ f (x) = ???

df (x) = ′f (x)dx∫∫ = f (x)+C

Έστω συνάρτηση f(x) µε πεδίο ορισµού D. Μία συνάρτηση F(x) καλείται αντιπαράγωγος εάν

F΄(x)= f(x)

για κάθε x στο D. Το σύνολο όλων των αντιπαραγώγων της f(x) είναι το αόριστο ολοκλήρωµα της συνάρτησης f(x) ως προς x , και συµβολίζεται µε

( )f x dx∫18/10/2016 ©2016 Γιάννης Σαριδάκης/ΕΕΜΗΥ/Πολυτεχνείο Κρήτης 20

( ) ( )f x dx F x C= +∫

( )

( )

1 1

11 1

1 1

cos cossin sin

sin sincos cos

n nn nx d xx dx C n x

n dx nddx x C xdx

kx d kxkxdx C kxk dx kkx d kxkxdx C kxk dx k

+ +⎛ ⎞= + ≠ − =⎜ ⎟+ +⎝ ⎠

= + =

⎛ ⎞= − + − =⎜ ⎟⎝ ⎠⎛ ⎞= + =⎜ ⎟⎝ ⎠

∫

∫

∫

∫18/10/2016 ©2016 Γιάννης Σαριδάκης/ΕΕΜΗΥ/Πολυτεχνείο Κρήτης 21

( )

( )

( )

( )

2 2

2 2

sec tan tan sec

csc cot cot csc

sec tan sec sec sec tan

csc cot csc csc csc cot

dxdx x C x xdxdxdx x C x xdxdx xdx x C x x xdxdx xdx x C x x xdx

= + =

= − + − =

= + =

= − + − =

∫

∫

∫

∫


( ) ( )f x dx F x C= +∫

( ) ( ) ( )kf x dx kF x C k f x dx= + =∫ ∫

[ ] [ ]( ) ( ) ( ) ( ) ( ) ( )f x g x dx F x G x C f x dx g x dx± = ± + = ±∫ ∫ ∫

( ) ( ) ( )f x dx F x C f x dx− = − + = −∫ ∫



v v = V max sin 12011"1

o

FIGURE 5.23 The graph of tire voltage V over a full cycle. Its average value over a half-cycle is 2V-!1T. Its average value over a full cycle is zero (Example 9).

5.5 Indefinite Integrals and the Substitution Method 289

SoLution 2: Substitute u = V'z2 + I instead.

Let u = \¥z2+1, J 2zdz = J 3u2du V'z2 + I u "3 = Z2 + 1, 3u 2 du = 2zdz.

= 3JUdU

U2 = 3'- + C 2 Integrate.

= 1(z2 + 1),/3 + C Replace uby (Z2 + 1)1/3. • 2

The Integrals of sin2 x and cos2 X

Sometimes we can use trigonometric identities to transform integrals we do not know bow to evaluate into ones we can evaluate using the substitution rule.

EXAMPLE 8

(a) J sin2 x dx = J I - 2x dx

= tJ(1 -cos2x) dx

= Ix _lsin2x + C = '! _ sin2x + C 2 2 2 2 4

(b) JCOS2Xdx = J I + dx = I + Sin42x + C

sin2 x = 1 - cos2x 2

2 l+cos2x cosx= 2 • EXAMPLE 9 We can model the voltage in the electrical wiring of a typical home with the sine function

v = V lWlX sin 120,,1,

which expresses the voltage V in volts as a function of time 1 in seconds. The function runs through 60 cycles each second (its frequency is 60 hertz, or 60 Hz). The positive constant V llW< (''vee max") is the peak voltage.

The average value of Vover the half-cycle from 0 to 1/120 sec (see Figure 5.23) is

I 1.1/12• . Vav = (1/120) _ O. Vmax sm 120"tdt

[ I ]1/12.

= 120Vmax - 120" cos 120"t •

Vmax = ---.r [ -COS" + cos 0]

2Vmax " . The average value of the voltage over a full cycle is zero, as we can see from Figure 5.23. (Also see Exercise 64.) If we measured the voltage with a standard moving-coil gal-vanometer, the meter would read zero.

To measure the voltage effectively, we use an instrument that measures the square root of the average value of the square of the voltage, namely

Vnns = V(V2).v'

v v = V max sin 12011"1

o

FIGURE 5.23 The graph of tire voltage V over a full cycle. Its average value over a half-cycle is 2V-!1T. Its average value over a full cycle is zero (Example 9).

5.5 Indefinite Integrals and the Substitution Method 289

SoLution 2: Substitute u = V'z2 + I instead.

Let u = \¥z2+1, J 2zdz = J 3u2du V'z2 + I u "3 = Z2 + 1, 3u 2 du = 2zdz.

= 3JUdU

U2 = 3'- + C 2 Integrate.

= 1(z2 + 1),/3 + C Replace uby (Z2 + 1)1/3. • 2

The Integrals of sin2 x and cos2 X

Sometimes we can use trigonometric identities to transform integrals we do not know bow to evaluate into ones we can evaluate using the substitution rule.

EXAMPLE 8

(a) J sin2 x dx = J I - 2x dx

= tJ(1 -cos2x) dx

= Ix _lsin2x + C = '! _ sin2x + C 2 2 2 2 4

(b) JCOS2Xdx = J I + dx = I + Sin42x + C

sin2 x = 1 - cos2x 2

2 l+cos2x cosx= 2 • EXAMPLE 9 We can model the voltage in the electrical wiring of a typical home with the sine function

v = V lWlX sin 120,,1,

which expresses the voltage V in volts as a function of time 1 in seconds. The function runs through 60 cycles each second (its frequency is 60 hertz, or 60 Hz). The positive constant V llW< (''vee max") is the peak voltage.

The average value of Vover the half-cycle from 0 to 1/120 sec (see Figure 5.23) is

I 1.1/12• . Vav = (1/120) _ O. Vmax sm 120"tdt

[ I ]1/12.

= 120Vmax - 120" cos 120"t •

Vmax = ---.r [ -COS" + cos 0]

2Vmax " . The average value of the voltage over a full cycle is zero, as we can see from Figure 5.23. (Also see Exercise 64.) If we measured the voltage with a standard moving-coil gal-vanometer, the meter would read zero.

To measure the voltage effectively, we use an instrument that measures the square root of the average value of the square of the voltage, namely

Vnns = V(V2).v'


236 Chapter 4: Applications of Derivatives

Exercises 4.7

finding Antlderlvatlves

This formula is more complicated than it needs to be. Ifwe combine C" -2C" and 5C, into a single arbitrnry constant C = C, - 2C, + 5C" the formula simplifies to

x' --x'+5x+C 3

and still gives all the possible antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate term-by-term. Write

J (x' - 2x + 5) dx = J x' dx - J 2x dx + J 5 dx

x' = 3 - x' + 5x + C.

Find the simplest antiderivative you can for each part and add the arbitrnry constant of integration at the end _

finding Indefinite Integrals Io Exercises 1-16, f"md an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

Io Exercises 17-54, find the most general antiderivative or indeflnite integral. Check your answers by differentiation.

1.L2x b.x' c.x'-2x+1 2. L 6x b. x7 c. x7 - 6x + 8

17.1 (x+ I)dx 18./(5-6x)dx 3. L -3x-4 b. x-4 c. x-4 + 2x + 3

-, 4. L 2x-' b. +x2 C. -x-3 + x-I 2

1 5 5. L 2 c. 2- 2 X x' x 2 b. _1_ , 1 6. L-- C. x -3 x' 2x' x

7. L b. _1_ c. Vx+ _1_ 2Vx Vx

b. _1_ 8. L 3 x c.

9 2 -1/3 b. 1x-2/3 c. -tx-4/3 . L}X 3 10 1 -1/2 b. _lx-'/2 c. . L 2x 2

19. 1 (3t2 + t) dt 20. 1 + 4t') dt 21. 1(2x' - 5x + 7)dx 22. l(i - x' - 3x')dx

23·/8,-x2-t)dx 24. 1 (t -:' + 2x) dx 25. lx-II' dx 26. lx-51' dx 27. 1 (Vx + dx 28. I(Yf +

29. 1(8Y - )I')dy 30. I(t- )I')dy 31. 12x(i - x -') dx 32. Ix-'(x+ l)dx

11. L -w sin '11'X b. 3 sinx c. sin 71'X - 3 sin 3x

12. L 'T1' cos '7TX b. 1rcos 'T1'X 'If X 2 2 c. COST + wooSX'

13. L see'x 2 e' x -see' 3x b. -se - c. 3 3 2

14. L ese'x b. -lese' 3x c. 1-8ese'2x 2 2 15. L cscxcotx b. -esc 5x cot 5x c. -'1r esc 1TX cot 1TX

2 2

33. 1 tv,;t; V,dt 34. 1 4 dt

35. 1(-2 cost) dt 36. 1(-5 sint) dt

37. dO 38. 13 cos 50 dO

39·/(-3eSC'x)dx 40. 16. L secxtanx b. 4 sec 3x tan 3x '1fX 'If X

c. secTtanT 41. 1 esc °2eot 0 dO 42. 1

43. 1(4secxtmx - 2sec'x)dx 44.1 k(csc'x - cscxcotx)dx

45. 1 (sin 2x - esc' x) dx 46. 1(2 cos 2x - 3 sin 3x) dx

47. 1 1 + dt

49. l(i + tm'B) dB 50./(2 + tm'B) dB

(Hint: 1 + tm' B = sec' B)

51. 1 cot?-xdx 52.1(1 - cot?-x)dx

(Hint: 1 + cot?- x = esc' x)

53. 1 cosB(tmB + sec B) dB 54. 1 cscB. dB cscO - 8m8

Checking Antiderivative Formulas VerifY the formulas in Exercises 55-j)O by differentiation.

1 ' (7x - 2)4 55. (7x - 2) dx = 28 + C

(3x + 5)-1 3 + C

57.1 sec'(5x - I)dx = ttm(5x - I) + C

58. 1 esc' (x I) dx = -3 cot (x I) + C

1 I I 59. (x + I)' dx = - x + I + C

1 I x 60. (x+I),dx=x+I+C

61. Right, or wrong? Say which for each formula and give a brief rea-son for each answer.

a. IXSinXdx = sinx + C

b. IXSinXdx=-XCOSX+C

c. IXSinXdx = -xcosx + sinx + C


a. 1 tmBsec'BdB = + C

b. 1 tmBsec'BdB = ktm'B + C

c. 1 tmBsec'BdB = ksec'B + C


1 (2x + I)' a. (2x + I)' dx = 3 + C

4.7 Antiderivatives 237

b. 13(2x + I)' dx = (2x + I)' + C

c. 16(2x + I)' dx = (2x + I)' + C

64. Right, or wrong? Say which for each formula and give a brief rea-son for each answer .

•. Yx'+x+ C

b.

c.

65. Right, or wrong? Give a brief reason why.

1 _-.::.:15",(x,--+---,:"3)"--' (x + 3)' dx= -- +C (X_2)4 x-2

66. Right, or wrong? Give a brief reason why.

1 _x_co_s_<,-x-,')-c-c-s_in_<,-x-,-') _ sin (x') , dx- x +C

x

Initial Yalue Problems 67. Which of the following graphs shows the solution of the initial

value problem

dy dx=2x, y=4whenx=l?

y

4

3

2

(1,4)

-1 0 1

(a)

y

-1 0

(b)

Give reasons for your answer.

(c)

68. Which of the following graphs shows the solution of the initial value problem

dy dx = -x, Y = I whenx = -I?

y y y

o x x Hy

o x

(0) (b) (c)

Give reasons for your answer.

0 0( ) , ( )dy f x y x ydx

= =

dydx

dx∫ = f (x)∫ dx

!y = F(x)+C , C = y0 − F(x0 )



238 Chapter 4: Applications of Derivatives

Solve the initial value problems in Exercises 69-1!8. dy

69. dx = 2x - 7, y(2) = 0

dy 70. dx = 10 - x, y(O) = -I

dy 1 71. dx = x 2 + X, X > 0; y(2) = I

dy 72. dx = 9x2 - 4x + 5, y(-I) = 0

73 dy = 3 -2/3 • dx x ,

dy I 74. dx = • " 2vx

y(-I) = -5

y(4) = 0

dx 75. dl = I + COSI, s(O) = 4

76. : = cos I + sinl, s(".) = I

77. = -".sin".O, r(O) = 0

dr 78. dO = cos,,-o, r(O) = 1

dv I 79. dl = Zsecltanl, v(O) = I

d'y 81. -2 = 2 - 6>:; y'(O) = 4, y(O) = I

dx

d'y 82. -2 = 0; y'(O) = 2, y(O) = 0

dx

d 2r 2 drl 83. dl2 = -I'; - = 1 r(l) = I dt 1=1 '

84 d2s = 31. tis I = 3 s(4) = 4

'd12 8' '

d'y 85. dx' = 6; y"(O) = -8, y'(O) = 0, y(O) = 5

d'O I. r.; 86. dl' = 0; 0"(0) = -2, 0'(0) = -Z' 0(0) = v2

87. y(4) = -sin t + cos t; = 7, y"(O) = y'(O) = -I, y(O) = 0

88. y(4) = -cosx + 8 sin2x; = 0, y"(O) = y'(O) = I, y(O) = 3

89. Find the curve y = f(x) in the xy-pIane that passes through the point (9, 4) and whose slope at each point is 3 Vx.

90. Uniquene •• of solution. If differentiable functions y = F(x) and y = G(x) both solve the initial value problem

dy dx = f(x), y(xo) = Yo.

on an interval I, most F(x) = G(x) for every x in n Give reasons for your answer.

Solution (Integral) Curves Exercises 91-94 show solution curves of differential eqoations. 10 each exercise, fmd an equation for the curve through the labeled point

91. y dJ.. 1 _ .! ,113 92. dx 3

93. dJ.. sin x - cos> dx y

Applications

94. y

95. Finding displacement from an antiderivati"" of ""Iocity a. Suppose that the velocity of a body moving along the s-axis is

tis dl v = 9.81 - 3.

i) Find the body's displacement aver the time interval from I = I tol 3giventhats = 5 when I O.

ii) Find the body's displacement from I = I to I 3 giveo thats = -2 when 1= O.

iii) Now find the body's displacement from I = 1 to I 3 given that s = So when I = O.

b. Suppose that the position s of a body moving along a coordinate line is a differentiable function of time t. Is it true that once you know an antiderivative of the velocity function tisldlYou can fmd the body's displacement from I = a to I = b even if you do not know the body's exact position at either of those times? Give reasons for your answer.

96. Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 mjsec'. How fast will the rocket be going I min later?

97. Stopping a cor in time You are driving along a highway at a steady 60 mph (88 fi/sec) when you see an accident shead and slan3 on the brakes. What constant deceleration is required to stop your car in 242 fi? To fmd out, carry out the following steps.

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