BÀI 3: PHÉP TÍNH TÍCH PHÂN

B?i 3: Ph?p t?nh t?ch ph?n

B?I 3: PH?P T?NH T?CH PH?N

Mc ti?u

Nm c c?c kh?i nim v t?ch ph?n bt nh, t?ch ph?n x?c nh, t?ch ph?n suy rng.

L?m c b?i tp v t?ch ph?n bt nh, t?ch ph?n x?c nh.

?p dng phn mm Maple t?nh t?ch ph?n.

Thi lng

Ni dung

Bn n?n d?nh mi tun khong 90 ph?t c k l? thuyt v? khong 120 ph?t trong v?ng hai tun l?m b?i tp nm vng ni dung b?i hc n?y.

B?i n?y gii thiu vi c?c bn c?c kh?i nim t?ch ph?n bt nh, t?ch ph?n x?c nh, t?ch ph?n suy rng v? c?c phng ph?p t?nh c?c loi t?ch ph?n n?y.

Ph?p t?nh t?ch ph?n l? mt trong hai ph?p t?nh c bn ca gii t?ch, c? nhiu ng dng trong b?i to?n k thut, kinh t...

Hng dn hc

Bn n?n c k l? thuyt nm c c?c kh?i nim t?ch ph?n bt nh, t?ch ph?n x?c nh v? c?c loi t?ch ph?n suy rng.

Bn n?n l?m c?ng nhiu b?i tp c?ng tt th?nh tho phung ph?p t?nh c?c loi t?ch ph?n ?.

MAT101_B?i 3_v2.3013101225

43

{{

3.1.

T?ch ph?n bt nh

B?i 3: Ph?p t?nh t?ch ph?n

3.1.1. Kh?i nim v t?ch ph?n bt nh

3.1.1.1. Nguy?n h?m

B?i n?y tr?nh b?y v ph?p t?nh t?ch ph?n, ?y l? ph?p to?n ngc ca ph?p t?nh o h?m (vi ph?n) ca h?m s. Nu ta cho trc mt h?m s f (x) th? c? tn ti hay kh?ng mt h?m s F(x) c? o h?m bng f (x) ? Nu tn ti, h?y t?m tt c c?c h?m s F(x) nh vy.

nh ngha: H?m s F(x) c gi l? mt nguy?n h?m ca h?m s f (x) tr?n mt khong D nu:

F '(x) f (x),x D , hay dF(x) f (x)dx .

V? d 1: V?: (sin x) ' cos x,x n?n sin x l? nguy?n h?m ca h?m s cos x tr?n .

V?:

arctg

x

1 1 x2

'

1 1 x2

2x (1 x2 )2

, x

1

n?n:

arctg x

1 1 x2

l?

mt

nguy?n

h?m

ca

h?m

s

1 1 x2

2x (1 x2 )2

tr?n \ 1.

nh l? sau ?y n?i rng nguy?n h?m ca mt h?m s cho trc kh?ng phi l? duy nht, nu bit mt nguy?n h?m th? ta c? th mi?u t c tt c c?c nguy?n h?m kh?c ca h?m s ?.

nh l?: Nu F(x) l? mt nguy?n h?m ca h?m s f (x) tr?n khong D th?:

H?m s F(x) C cng l? mt nguy?n h?m ca h?m s f (x) , vi C l? mt hng s bt k. Ngc li, mi nguy?n h?m ca h?m s f (x) u vit c di dng F(x) C , trong ? C l? mt hng s.

Chng minh: Gi s C l? mt hng s bt k, ta c?:

F(x) C ' F '(x) f (x) vi mi x D .

Theo nh ngha F(x) C cng l? mt nguy?n h?m ca h?m s f (x) tr?n khong D. Ngc li, gi s (x) l? mt nguy?n h?m bt k ca h?m s f (x) tr?n khong D. Ta c?:

F(x) (x)' F'(x) '(x) f (x) f (x) 0,x D .

Suy ra F(x) (x) nhn gi? tr hng s tr?n khong D:

F(x) (x) C (x) F(x) C,x D .

Nh vy biu thc F(x) C biu din tt c c?c nguy?n h?m ca h?m s f (x) , mi hng s C tng ng cho ta mt nguy?n h?m.

44

MAT101_B?i 3_v2.3013101225

3.1.1.2. T?ch ph?n bt nh

B?i 3: Ph?p t?nh t?ch ph?n

nh ngha: T?ch ph?n bt nh ca mt h?m s f (x) l? h c?c nguy?n h?m F(x) C ; vi x D ; trong ? F(x) l? mt nguy?n h?m ca h?m s f (x) v? C l? mt hng s bt k. T?ch

ph?n bt nh ca f (x)dx c k? hiu l?: f (x)dx .

Biu thc f (x)dx c gi l? biu thc di du t?ch ph?n v? h?m s f c gi l? h?m s di du t?ch ph?n.

Vy: f (x)dx F(x) C , vi F(x) l? nguy?n h?m ca f (x) .

V? d 2:

cos xdx sin x C

exdx ex C .

3.1.1.3. C?c t?nh cht c bn ca t?ch ph?n x?c nh

f (x)dx ' f (x) hay d f (x)dx f (x)dx

F'(x)dx F(x) C hay dF(x) F(x) C

af (x)dx a f (x)dx , ( a l? hng s kh?c 0)

f (x) g(x)dx f (x)dx g(x)dx .

Hai t?nh cht cui c?ng l? t?nh cht tuyn t?nh ca t?ch ph?n bt nh, ta c? th vit chung:

f (x) g(x)dx f (x)dx g(x)dx

trong ? , l? c?c hng s kh?ng ng thi bng 0

C?c t?nh cht n?i tr?n c chng minh trc tip t nh ngha ca t?ch ph?n bt nh.

3.1.1.4. C?c c?ng thc t?ch ph?n c bn

C?c c?ng thc t?ch ph?n sau ?y c chng minh bng nh ngha:

xdx

x 1 1

C,

(

1)

sin xdx cos x C

dx sin2 x

cotg

x

C

axdx

ax ln a

C,

(a

0,

a

1)

dx

a2 x2

1 2a

ln

ax ax

C

dx ln x x2 C

x2

dx x

ln

x

C

cos xdx sin x C

dx cos2

x

tg

x

C

exdx ex C

dx x2 a2

1 a

arctg

x a

C

dx a2 x2

arcsin

x a

C

MAT101_B?i 3_v1.0013101225

45

{{

3.1.2. C?c phng ph?p t?nh t?ch ph?n bt nh

B?i 3: Ph?p t?nh t?ch ph?n

3.1.2.1. Phng ph?p khai trin

t?nh mt t?ch ph?n bt k, ta cn s dng c?c phng ph?p th?ch hp a v c?c t?ch ph?n ? c? trong bng c?c c?ng thc t?ch ph?n c bn tr?n. Mt phng ph?p n gin l? phng ph?p khai trin. Phng ph?p n?y da tr?n t?nh cht tuyn t?nh ca t?ch ph?n bt nh:

f (x) g(x)dx f (x)dx g(x)dx .

Ta ph?n t?ch h?m s di du t?ch ph?n th?nh tng (hiu) ca c?c h?m s n gin m? ? bit c nguy?n h?m ca ch?ng, c?c hng s c a ra b?n ngo?i du t?ch ph?n.

V? d 3:

(2x

x 3x2 )dx 2

3

x 2dx 3

x 2 dx

4 5

5

x2

x3

C

2sin

x

x3

1 x

dx

2sin

xdx

x3dx

dx x

2cos

x

x4 4

ln

x

C

dx x2 (1 x2 )

1 x2

1 1 x2

dx

1 x

arctg

x

C

.

3.1.2.2. Phng ph?p bin i biu thc vi ph?n

Nhn x?t:

Nu: f (x)dx F(x) C th? f (u)du F(u) C ; trong ? u u(x) l? mt h?m s

kh vi li?n tc. Ta c? th kim tra li bng c?ch o h?m hai v theo x.

S dng t?nh cht n?y, ta bin i biu thc di du t?ch ph?n g(x)dx v dng:

g(x)dx f (u(x))u '(x)dx

trong ? f (x) l? mt h?m s m? ta d d?ng t?m c nguy?n h?m F(x) . Khi ? t?ch ph?n cn t?nh tr th?nh:

g(x)dx f (u(x))u '(x)dx f (u(x))du F(u(x)) C a 0

Trong trng hp n gin u(x) ax b th? du adx , do ? nu

f (x)dx

F(x) C

ta suy ra: f (ax b)dx

1 a

F(ax

b)

C

a 0

V? d 4:

sin

axdx

1 a

cos

ax

C

.

a

0

eaxdx

eax a

C

a

0

esin x cos xdx esin xd(sin x) esin x C

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MAT101_B?i 3_v2.3013101225

B?i 3: Ph?p t?nh t?ch ph?n

dx cos4

x

(1

tg2

x)d(tg

x)

tg3 3

x

tg

x

C

x

1

3x 2

dx

1 6

1

3x

2

d(1

3x

2

)

1 9

3

1 3x2 C

I

arccos x arcsin 1 x2

x

dx

2

arcsin

x

arcsin

xd(arcsin

x)

I

4

arcsin 2

x

1 3

arcsin3

x

C

.

3.1.2.3. Phng ph?p i bin

X?t t?ch ph?n I f (x)dx ; trong ? f (x) l? mt h?m s li?n tc. t?nh t?ch ph?n

n?y, ta t?m c?ch chuyn sang t?nh t?ch ph?n kh?c ca mt h?m s kh?c bng mt ph?p i bin sao cho biu thc di du t?ch ph?n i vi bin t c? th t?m c nguy?n h?m mt c?ch n gin hn. Ta chia phng ph?p i bin l?m hai trng hp l? i bin xu?i x (t) v? i bin ngc t (x) .

Ph?p i bin th nht: t x (t) ; trong ? (t) l? mt h?m s n iu, v? c? o h?m li?n tc. Khi ? ta c?:

I f (x)dx f (t) '(t)dt

Gi s h?m s g(t) f (t) '(t) c? nguy?n h?m l? h?m G(t) , v? t h(x) l?

h?m s ngc ca h?m s x (t) , ta c?:

I g(t)dt G(t) C I G h(x) C .

Ph?p i bin th hai:

t t (x) , trong ? (x) l? mt h?m s c? o h?m li?n tc, v? ta vit c

h?m f (x) g (x) '(x) . Khi ? ta c?:

CH? ? :

I f (x)dx g(x) '(x)dx .

Gi s h?m s g(t) c? nguy?n h?m l? h?m s G(t) , ta c?:

I G(x) C .

Khi t?nh t?ch ph?n bt nh bng phng ph?p i bin s, sau khi t?m c nguy?n h?m theo bin s mi, phi i li th?nh h?m s ca bin s c.

V? d 5:

a) T?nh t?ch ph?n: I1

x dx 2x

t

x

2 sin 2

t,

t

0,

2

,

ta

t?nh

c:

dx 4sin t cos tdt ;

MAT101_B?i 3_v1.0013101225

47

B?i 3: Ph?p t?nh t?ch ph?n

{{

x 2x

2 sin2 t 2(1 sin2

t)

tg t .

Suy ra: I1

2

x

x

dx

4

sin 2

tdt

2t

sin

2t

C

.

i li bin x, vi t arcsin

x 2

,

ta

thu

c:

I1

2

x

x

dx

2

arcsin

x 2

2x x2 C.

b) T?nh t?ch ph?n I2

e2x ex

1

dx

.

t ex t exdx dt , ta c?:

I2

t

t

1

dt

1

t

1

1

dt

t

ln

t

1

C

.

i li bin x, ta c: I2 ex ln(ex 1) C .

c) T?nh t?ch ph?n I3

dx . 1 4x

t t 2x dt 2x ln 2dx , t?ch ph?n tr th?nh:

I3

dt t ln 2 1 t2

1 ln 2

dt t2 1

1 ln 2

ln(t

t2 1) C .

i

li

bin

x,

ta

c?:

I3

1 ln 2

ln(2

x

4x 1) C .

3.1.2.4. Phng ph?p t?ch ph?n tng phn

Gi s u u(x) v? v v(x) l? c?c h?m s c? o h?m li?n tc. Theo quy tc ly vi ph?n

d(uv) udv vdu uv d(uv) udv vdu . Suy ra : udv uv vdu . X?t t?ch ph?n: I f (x)dx .

Ta cn biu din:

f (x)dx g(x)h(x)dx g(x)h(x)dx udv

v? ?p dng c?ng thc t?ch ph?n tng phn vi c?c h?m s u g(x); v h(x)dx .

Ta thng s dng phng ph?p n?y khi biu thc di du t?ch ph?n cha mt trong c?c h?m s sau ?y: ln x;ax ; h?m s lng gi?c, h?m s lng gi?c ngc. C th:

Trong c?c t?ch ph?n xnekxdx; xn sin kxdx; xn cos kxdx , n nguy?n dng, ta thng

chn: u xn

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MAT101_B?i 3_v2.3013101225

B?i 3: Ph?p t?nh t?ch ph?n

Trong c?c t?ch ph?n x lnn xdx , 1 v? n nguy?n dng, ta thng chn u lnn x Trong t?ch ph?n xn arctg kxdx; xn arcsin kxdx , n nguy?n dng, ta thng chn:

u arctg kx hoc u arcsin kx ; dv xndx . V? d 6: T?nh c?c t?ch ph?n bt nh:

a) I1 ln xdx x ln x dx x ln x x C . b) I2 x2 sin xdx .

t u x2, dv sin xdx v cos x , ta c:

I2 x2 cos x 2 x cos xdx .

t u x, dv cos xdx v sin x , ta c:

I2 x2 cos x 2 x sin x sin xdx x2 cos x 2xsin x 2cos x C.

c)

I3

xexdx (x 1)2

.

t

u

xex ; dv

dx (x 1)2

v

x

1 1

;

du

(x 1)exdx

,

ta

c:

I3

xex x 1

e x dx

xex x 1

ex

C

ex x 1

C

.

d) I4

xexdx . 1 ex

t 1 ex t exdx 2dt ; ta c?: 1 ex

I4 2ln(t 1) ln(t 1)dt 2(t 1)ln(t 1) 2(t 1)ln(t 1) 4t C .

i li bin x ta c?:

xexdx 2(x 2) 1 ex 4 ln 1 1 ex 2x C . 1 ex

e)

I5

x arcsin x dx . 1 x2

MAT101_B?i 3_v1.0013101225

49

B?i 3: Ph?p t?nh t?ch ph?n

{{

t u arcsin x;dv xdx du dx ; v 1 x2 , ta c:

1 x2

1 x2

I5 1 x2 arcsin x dx 1 x2 arcsin x x C . f) I6 ex cos 2xdx .

t u cos 2x;dv exdx v ex ;du 2sin 2xdx ; ta c:

3.1.3.

I6 ex cos 2x 2 ex sin 2xdx .

t u sin 2x;dv exdx v ex ;du 2 cos 2xdx ; ta c:

I6 ex cos2x 2 ex sin2x 2 ex cos2xdx ex cos2x 2ex sin2x 4I6 5C.

Vy:

I6

ex 5

cos

2x

2 sin

2x

C

.

Trong c?c mc sau ?y ch?ng ta s x?t t?ch ph?n bt nh ca mt s dng h?m c bn: H?m ph?n thc hu t, h?m lng gi?c, h?m cha cn thc v? tr?nh b?y mt s phng ph?p gii chung i vi t?ch ph?n c?c h?m n?y.

T?ch ph?n h?m ph?n thc hu t

nh ngha:

Mt

h?m

ph?n

thc

hu

t

l?

mt

h?m

s

c?

dng:

f (x)

P(x) Q(x)

,

trong ? P(x), Q(x) l? c?c a thc ca x.

Mt ph?n thc hu t c? bc ca a thc t s nh hn bc ca a thc mu s l? mt ph?n thc hu t thc s.

Bng ph?p chia a thc, chia P(x) cho Q(x) ta lu?n a c mt h?m ph?n thc

hu t v dng:

f

(x)

H(x)

r(x) Q(x)

Trong ? H(x) l? a thc thng, r(x) l? phn d trong ph?p chia.

Khi ?

r(x) Q(x)

l? mt ph?n thc hu t thc s. Nguy?n h?m ca a thc

H(x)

c

t?m bi c?ng thc t?ch ph?n c bn:

xndx

x n1 n 1

C

;

n

nguy?n

dng.

Ta s x?t vic t?m nguy?n h?m ca ph?n thc hu t c?n li

r(x) Q(x)

trong hai trng hp

c bit: Mu s ca ph?n thc l? a thc bc nht hoc a thc bc hai. Trong nhng

trng hp mu s phc tp hn, ch?ng ta s dng phng ph?p h s bt nh a v hai trng hp tr?n.

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MAT101_B?i 3_v2.3013101225

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