(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (

Similar documents
18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

1

r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a

y = f(x) y = f( + h) f(), x = h dy dx f () f (derivtive) (differentition) (velocity) p(t) =(x(t),y(t),z(t)) ( dp dx dt = dt, dy dt, dz ) dt f () > f x

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

Chap11.dvi

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =

2010 II / y = e x y = log x = log e x 2. ( e x ) = e x 3. ( ) log x = 1 x 1.2 Warming Up 1 u = log a M a u = M a 0

() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.

i

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2

2012 IA 8 I p.3, 2 p.19, 3 p.19, 4 p.22, 5 p.27, 6 p.27, 7 p

- II

, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x

body.dvi

x i [, b], (i 0, 1, 2,, n),, [, b], [, b] [x 0, x 1 ] [x 1, x 2 ] [x n 1, x n ] ( 2 ). x 0 x 1 x 2 x 3 x n 1 x n b 2: [, b].,, (1) x 0, x 1, x 2,, x n

2014 S hara/lectures/lectures-j.html r 1 S phone: ,

名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト

ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)


<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

DVIOUT

1 R n (x (k) = (x (k) 1,, x(k) n )) k 1 lim k,l x(k) x (l) = 0 (x (k) ) 1.1. (i) R n U U, r > 0, r () U (ii) R n F F F (iii) R n S S S = { R n ; r > 0

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (

III ϵ-n ϵ-n lim n a n = α n a n α 1 lim a n = 0 1 n a k n n k= ϵ-n 1.1

f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f

17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,

ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.

A S hara/lectures/lectures-j.html ϵ-n 1 ϵ-n lim n a n = α n a n α 2 lim a n = 0 1 n a k n n k= ϵ

29

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)

..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A

[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t )

< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3)

Riemann-Stieltjes Poland S. Lojasiewicz [1] An introduction to the theory of real functions, John Wiley & Sons, Ltd., Chichester, 1988.,,,,. Riemann-S


M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

2 p T, Q

JAふじかわNo43_ indd

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10

(1) 1 y = 2 = = b (2) 2 y = 2 = 2 = 2 + h B h h h< h 2 h

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29

.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,

Acrobat Distiller, Job 128


訪問看護ステーションにおける安全性及び安定的なサービス提供の確保に関する調査研究事業報告書

(2000 )

(1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9

i 18 2H 2 + O 2 2H 2 + ( ) 3K

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x

2016.

第121回関東連合産科婦人科学会総会・学術集会 プログラム・抄録


O f(x) x = A = lim h f( + h) f() h A (differentil coefficient) f f () y = f(x) y = f( + h) f(), x = h dy dx f () f (derivtive) (differentition) * t (v

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (

6.1 (P (P (P (P (P (P (, P (, P.

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

lecture

A

No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y


6. Euler x

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +

6.1 (P (P (P (P (P (P (, P (, P.101

1 29 ( ) I II III A B (120 ) 2 5 I II III A B (120 ) 1, 6 8 I II A B (120 ) 1, 6, 7 I II A B (100 ) 1 OAB A B OA = 2 OA OB = 3 OB A B 2 :

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

v er.1/ c /(21)

II III I ~ 2 ~

中堅中小企業向け秘密保持マニュアル


PR映画-1

- 2 -



1 (1) (2)

Microsoft Word - 表紙.docx


( )/2 hara/lectures/lectures-j.html 2, {H} {T } S = {H, T } {(H, H), (H, T )} {(H, T ), (T, T )} {(H, H), (T, T )} {1

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

1 (1) () (3) I 0 3 I I d θ = L () dt θ L L θ I d θ = L = κθ (3) dt κ T I T = π κ (4) T I κ κ κ L l a θ L r δr δl L θ ϕ ϕ = rθ (5) l

DE-resume

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s

1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1

f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) B p.1/14

1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =

simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =

newmain.dvi

I y = f(x) a I a x I x = a + x 1 f(x) f(a) x a = f(a + x) f(a) x (11.1) x a x 0 f(x) f(a) f(a + x) f(a) lim = lim x a x a x 0 x (11.2) f(x) x

Transcription:

B 4 4 4 52 4/ 9/ 3/3 6 9.. y = x 2 x x =

(, ) (, ) S = 2 = 2 4 4 [, ] 4 4 4 ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, 4 4 4 4 4 k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) 2 2 + ( ) 3 2 + ( 4 4 4 4 4 4 4 4 4 ( ( = ) 2 ( ) 2 2 ( ) 3 2 ( ) ) 4 2 + + + 4 4 4 4 4 = 4 ( ) k 2 4 4 = 4 k= 6 = 4 6 4 k= 4 6 k 2 = 5 32 =.46 (4 + )(2 4 + ) n ( 2 + 2 2 + + n 2 ) = k 2 = n (n + )(2n + ) 6 k= ) 2 2

8 4 S S 4 8 2. [, ] 8 S 8. 2. 3. S 8 n n 3. [, ] n S n. 2. 3. S n 3

n n S n lim S n n lim S n = lim n n = lim n (n + )(2n + ) 6n2 ( + ) ( 2 + 6 n n = 6 2 = 3 ) lim n S n 3 [, ] 3 y = x2 x x = 3 S n lim S n n 3 [, ] 4 4 4 ( ) k 2 k =, 2, 3 4 s 4 s 4 = 4 3 k= ( ) k 2 = 4 4 6 3 k= k 2 = 7 32 4

4. [, ] 8 s 8. 2. 3. s 8 5. [, ] n s n. 2. 3. s n n s n lim n s n lim s n = lim n n = lim n (n )(2n ) 6n2 ( ) ( 2 6 n n = 6 2 = 3 [, ] 3 n s n < S < S n lim s n = lim n S n = 3 y = x 2 x x = S 3 S x 2 dx ) 2 y = x 2 f(x) y = f(x) x x = x = b < b 5

f(x) =x x x 2 x 3... x n =b [, b] n = x, x, x 2,..., x n, x n = b x k x k (k =, 2,..., n) f(x k ) (k =, 2,..., n) n n S n =f(x )(x x ) + f(x 2 )(x 2 x ) + + f(x n )(x n x n ) n = f(x k )(x k x k ) k= S = lim n k= S b n f(x k )(x k x k ) b f(x) dx f(x) dx f b integrl n lim f(x k ) (x k x k ) n k= b f(x) S sum dx S 6

f(x) f(x) dx n ( ) f(x) = x 2 S = x 2 k 2 dx lim n n n k= n lim f(x k )(x k x k ) b n k=. 2. 3. b b f(x) dx = f(x) dx = c f(x) dx = f(x) dx+ b f(x) dx b c f(x) dx 2 b b f(x) f(x) f(x) x x f(x) dx f(x) f(x) dx f(x) f(x) F (x) f(x) F (x) = f(x) b f(x) dx = F (b) F () 2. F (x) = x f(x) dx f(x) 2. f(x) F (x) b f(x) dx = F (b) F ().2 f(x) F (x) = x F (x) = f(x) 7 f(x) dx

6. f(x) = x [, 5] F (x) =. F (), F (), F (2), F (3), F (4), F (5) 2. x 5 F (x) = 3. F (x) F (x) = x 7. f(x) = 2x [, 5] F (x) = x. F (), F (), F (2), F (3), F (4), F (5) 2. x 5 F (x) = 3. F (x) F (x) = x x dx x dx = x 2x dx 2x dx = 8. f(x) = 2 x [, 5] F (x) = x. F (), F (), F (2), F (3), F (4), F (5) 2. x 5 F (x) = 3. F (x) F (x) = x 2 2 x dx = ( x 5) x x ( x 5) 2x x x dx x dx ( x 5) 2 x 9 (). [, b] f x b F (x) = x f(x) dx F F (x) = f(x) f(x) F (x) x f(t) dt F (x) 2x dx x 2 x dx 8

Proof. F (x) = x f(x) dx F (x) x f(x) f(x) S = F (x) S x x+h x f(x) dx x+h S f(x) F (x) S = F (x + h) F (x) S F (x) h f(x) h f(x + h) f(x) h < S < f(x + h) h h h > f(x) < S h < f(x + h) S = F (x + h) F (x) h f(x) < F (x + h) F (x) h < f(x + h) F (x + h) F (x) lim f(x) < lim < lim f(x + h) h h h h F (x + h) F (x) lim f(x) = f(x) lim h h h F (x) = f(x) = F (x) lim h f(x + h) = f(x) Remrk. x f(x) dx x x f(x)dx x x 2 x t x f(t) dt x x b t t x 9

.3 f(x). f(x) f(x) F (x) = f(x) F (x) f(x) 2 ( 2). F (x) f(x) b f(x) dx = F (b) F () Proof. G(x) = x f(x) dx 9G (x) = f(x) F (x) f(x) F (x) = f(x) G(x) F (x) (G(x) F (x)) = G (x) F (x) = f(x) f(x) = C x G() = x b G(x) F (x) = C G() F () = C f(x) dx = C = F () b G(x) F (x) = F () f(x) dx = G(b) = F (b) F () F (b) F () [ ] b F (x) F (x) b b [ ] b f(t) dt = F (x)

3.. 2. x 2 dx x 2 F (x) x 2 [ ] F (x) F () F () 2 x 2 dx f(x).4 ( ) 3 x3 = x 2 ( ) 3 x3 x 2 3 x3 + 3 = x 2 ( 3 x3 + 3 x 2 3 x3 + ) 2 = x 2 3 x3 + 2 x 2 x 2 4. f(x) f(x) f(x)dx f(x) dx f(x) f(x) x 2 C 3 x3 + C x 2 x 2 dx = 3 x3 + C 5. f(x) F (x) f(x) C F (x) + C C = F (x) f(x)dx = F (x) + C C f f

f(x) f(x) dx x dx =? x x x dx x x dx? (?)? x (?) = x? 2 x2 x x x x ( ) 2 x2 + C x dx 2 x2 + C x dx = 2 x2 + C f(x).4. x n x dx x 2 dx x 2 x 2 dx (?)? x n dx x 2 dx =? (?) = x 2? = 3 x3 + C x 2 dx = 3 x3 + C x n dx =? (?) = x n? = n + xn+ + C 2

6. x n dx = n + xn+ + C (n =,, 2,... ) dx = x + C dx dx 7. C. x 4 dx 3. x 6 dx 5. x dx 2. x 5 dx 4. x 99 dx.4.2 8.. kf(x) dx = k f(x) dx k 2. 3. ( f(x) + g(x)) dx = ( f(x) g(x)) dx = f(x) dx + f(x) dx g(x) dx g(x) dx Proof. 2. ( f(x) + g(x) (f(x) + g(x)) dx (?)? f(x) dx + ( g(x) dx) = ( f(x) dx) + (f(x) + g(x)) dx =? g(x) dx) = f(x) + g(x) (?) = f(x) + g(x)? = f(x) dx + g(x) dx (f(x) + g(x)) dx = f(x) dx + g(x) dx 9. 3

2. (8x 3 5x + 3) dx = 8 = 8 x 3 dx 5 4 x4 5 x dx + 3 = 2x 4 5 2 x2 + 3x + C dx 2 x2 + 3x + C 2.. (3x 2 4x 2) dx 2. (2x + x 3 ) dx 3. (x )(2x 3) dx 4. t 3 (t 6) dt 5. (2y + 5) 2 dy 6. 5 dx 7. 7 dy 8. 9... 2. 3. 4. (2x 3 3x 2 ) dx (x 3)(2x + ) dx (3x + 2) dx (x 2 6x + 5) dx (8x 3 + 2x 2 + 4x) dx (x + 2)(x 2) dx (3x + 7)(x ) dx, 2, 3 22. 2 F (x). F (x) = x 2 4x, F (3) = 4 () F (x) x 2 4x F (x) = (x 2 4x) dx = (b) F (3) = 4 C C = (c) F (x) = 2. F (x) = 6x 2 4x 5, F (2) = 2 3. F (x) = 3 x, F () = 2 3 4. F (x) = 5x 2 2x + 6, F ( ) = 9 5. F (x) = (3x )( x), F () = 3 4,5 f(x) F (x) f(x) dx F (x) 4

23.. y = f(x) (, ) (x, f(x)) 3(x 2 ) f(x) 2. y = f(x) (, 2) (2, ) (x, f(x)) x 3 x f(x).4.3 x α 24. α x α dx = xα+ α + + C f(x) dx dx f(x) Remrk 25. α = α = x dx 26.. 2. 3. dx x 3 dx x 2 dx x 4. 5. 6. x 6 dx 3 x dx x x dx 7. 8. 6x 4 dx dx x n n n 2 27.. ( x 2 x 3 ) dx 3. (x x) 2 dx 2. ( x + x ) 2 dx 4. x 3 2x 2 dx x 5 5

.5 28. f(x) b F (x) b f(x) dx = 29. f(x) dx f(x) [ ] b F (x) = F (b) F (). 4 x 3 dx x 3 dx = F (x) F (4) F () 2. 3. 2 3 3 (y 2 2y 3)dy (x 3) 2 dx < 6

3.. 9 4 dx x. 3 2 x 2 dx 5. 3 (t 3) 2 dt 2. 3 x 2 dx 2. 5 2 3dx 6. 3 (y 3 4y)dy 3. 3 x 3 dx 3. (2x 2 + 3x + 5)dx 3. β α (x α)(x β) dx = 6 (β α)3 32..6 33. f(x) f( x) = f(x) f( x) = f(x) 34. f(x) f(x) f(x)dx = f(x)dx = 2 f(x)dx 3 4 5 6 y 7

35.. 2. 2 2 2 2 x 2 dx x 4 dx 3. 4. 3 3 (x 3 + 2x 2 + ) dx (4x 3 + 6x 2 9x ) dx 5. 3 3 (x 3) 2 dx 2 2. 36. [, b] f(x) y = f(x) x 2 x =, x = b S f(x) S = b f(x) S = f(x) dx b f(x) dx f(x) S = b f(x) dx f(x) + 37. y = x 3 3x 2 + 2x x S Proof.. x y = x 3 3x 2 +2x = x(x 2)(x ) y = x =,, 2 2. S = (x 3 3x 2 + 2x) dx 2 (x 3 3x 2 + 2x) dx = 2 38. x S 8

. y = 3x 2 x 3 2. y = x 3 + x 2 2x 3. y = x 2 5x + 4 3 4. y = x 2 x 6 3 39. x 2 x =, x = 4. y = 4x x 2 2. y = x 3 9

4. [, b] f(x) g(x) 2 y = f(x), y = g(x) 2 x =, x = b S S = b ( ) f(x) g(x) dx. g(x) b f(x) g(x) S = = b b f(x) dx b g(x) dx ( ) f(x) g(x) dx 2. f(x), g(x) y m f(x) + m g(x) + m f(x) g(x) S = = = b b b (f(x) + m) dx b (g(x) + m) dx ( ) f(x) + m (g(x) + m) dx ( ) f(x) g(x) dx b 4. y = x 2 + 2x + 4 y = x 2. 2 3 x 2 + 2x + 4 = x 2 2x 2 + 2x + 4 = 2(x + )(x 2) = 2. 3 S = 2 2(x + )(x 2) dx = 2 42. 2 (x + )(x 2) dx = 2 6 (2 + )3 = 9. y = x 2 4 y = x 2 2. y = 4x x 2 y = x 4 3. y = x 2 3x 4 y = 2 + x x 2 4. y = 2x 2 7x + 9 y = 5x x 2 2

43.. y = x 2 3x + 5 y = 3 3. y = 2x 2 +3x 5 y = x 2 +4x 3 2. y = 2x 2 + 3x 5 y = x 4. y = x 3 x 2 y = x 44.. y = x 2 x 2. y = x 3 x y 4. y = x 2 + 2x + 3 x 5. y = x 2 y = x 2 + 3x + 5 x = x = 2 3. y = x 2 x 3 x 45. x x =, x = 2. y = x 2 2x + 2 2. y = x 2 + 5x 6 2.2 46. 2 f(x) = x + b f(x) dx =, xf(x) dx = 47. 2 f(x) = x 2 + bx +. f () = 4, 2. f(2) = 3, f(x) dx = xf(x) dx = 3 n lim f(x k ) (x k x k ) n k= b f(x) dx 2 2

(km/ ).5 5 ( ) km/ 5.5km/.5km km/ 5 5.5km/ km/ km 5 5.5 5 = 7.5 5 7.5 km 5 5 f() 2 f(2) 3 f(3) 4 f(4) 5 f(5) = 7.5 (km/ ) 5.5 5 ( ) 5 f(x) dx km/h h = km kg/h h = kg /kg kg = / = / = / = 22

48. 2 f(x) = x 2 3x + 25. 2. 6 2 49. x kg kg y y = 5 4x + 6x 2 3 kg 7 kg 4 kg 5. 3 t y = 4 3 x 4 9 x2 5 5. kg 3 x kg 3x 2. x kg 2..4 kg 3. 2.3 kg 4. x kg 52. x kg kg /kg y = x 2 6x + 2 5 kg 23

[],, [2] 4, 5, IIIC+α [3] I II III,,, [4], [5], [6], S.,, [7], [8] ei =-,, ; (2/) [9], [], [], [2], [3], [4] mth stories,, (29//) [5]!!, [6] :,,, ; (23/3/9) [7],,,,, [8], A.C., K. [9] mth stories, [2], [2],, 24

[22], [23] 96, [24], [25],, (23/5) [26]!,, (2/7/6) [27],,, [28], [29],, [], [2], [3] [4], A.C., K. [5], [6], [7], [8], [9] 4 5 6, [], [] 3 3, [2], [3] :, 25

2.3 (f(g(x))) = f (g(x)) g (x) f(x + b) dx =? (?) = f(x + b)? f(x + b) f(x + b) dx (?)? F (t) f(t) F (t) = f(t) ( ) F (x + b) = F (x + b) = f(x + b)? = F (x + b) + C 53. F (t) f(t) f(x + b) dx = F (x + b) + C f(t) t x+b (x+b) α sin(x + b) 54. (2x ) dx Proof. t = 2x f(t) = t F (t) = (2x ) dx = 2 (2x ) + C = 22 (2x ) + C t dt = t + C f(x + b) 2 x + b t 26

55.. (2x + ) 2 2. (5 4x) 3 3. (x + 3) 2 4. (3x 2) 4 5. (4 3x) 3 6. 7. (2x + 3) 2 3x + 8. (3x 2) 2 9. ( 4x) 3. ( ) 2 2 3 x +. (x + 2) 2. ( 5x) 3 3. 6 ( 2x) 4 4. 4 x 5. 6. 7. 56. Proof. (2x ) dx (2x ) dx = [ 22 (2x ) 2x + 3 ( x 3 + 8 ) 3 8x + 7 ] = 57.. 2. 3. 4. 5. 6. 7. 3 2 4 2 3 2 2 5 (x 3) 2 dx (4 x) 3 dx (2x ) 3 dx (3 2x) dx (2x ) 3 dx 5 2x dx + 3x dx 8. 9... 2. 3. 2 3 2 2 2 5 2 3 2 (2x 3) 4 dx dx 3x 5 5 2x dx + 3x dx (2x 3) 4 dx dx 3x 5 27