Antrojo tipo kreivinis integralas fizikoje reiškia jėgų lauko darbą.
∫
A
B
P
(
x
,
y
)
d
x
+
Q
(
x
,
y
)
d
y
=
−
∫
B
A
P
(
x
,
y
)
d
x
+
Q
(
x
,
y
)
d
y
.
{\displaystyle \int _{AB}P(x,y)dx+Q(x,y)dy=-\int _{BA}P(x,y)dx+Q(x,y)dy.}
L :
∫
L
P
(
x
,
y
)
d
x
+
Q
(
x
,
y
)
d
y
+
R
(
x
,
y
)
d
z
.
{\displaystyle \int _{L}P(x,y)dx+Q(x,y)dy+R(x,y)dz.}
[ keisti ] Tarkime, kad kreivės L lanko AB parametrinės lygtis yra
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
A atitinka parametro t reikšme
t
0
{\displaystyle t_{0}}
B - reikšmė T . Dar sakykime, kad x(t), y(t) ir jų isvestinės x'(t), y'(t) yra dolydžios atkarpoje [
t
0
{\displaystyle t_{0}}
P
(
x
,
y
)
{\displaystyle P(x,y)}
Q
(
x
,
y
)
{\displaystyle Q(x,y)}
L taškuose funkcijos. Tuomet
∫
A
B
P
(
x
,
y
)
d
x
+
Q
(
x
,
y
)
d
y
=
∫
t
0
T
P
(
x
(
t
)
,
y
(
t
)
)
x
′
(
t
)
d
t
+
Q
(
x
(
t
)
,
y
(
t
)
)
y
′
(
t
)
d
t
.
{\displaystyle \int _{AB}P(x,y)dx+Q(x,y)dy=\int _{t_{0}}^{T}P(x(t),y(t))x'(t)dt+Q(x(t),y(t))y'(t)dt.}
∫
A
B
P
(
x
,
y
,
z
)
d
x
+
Q
(
x
,
y
,
z
)
d
y
+
R
(
x
,
y
,
z
)
d
z
=
{\displaystyle \int _{AB}P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz=}
=
∫
α
β
[
P
(
x
(
t
)
,
y
(
t
)
,
z
(
t
)
)
x
′
(
t
)
+
Q
(
x
(
t
)
,
y
(
t
)
,
z
(
t
)
)
y
′
(
t
)
+
R
(
x
(
t
)
,
y
(
t
)
,
z
(
t
)
)
z
′
(
t
)
]
d
t
.
{\displaystyle =\int _{\alpha }^{\beta }[P(x(t),y(t),z(t))x'(t)+Q(x(t),y(t),z(t))y'(t)+R(x(t),y(t),z(t))z'(t)]dt.}
Apskaičiuokime
∫
L
y
d
x
−
x
d
y
,
{\displaystyle \int _{L}ydx-xdy,}
L - apskritimo
x
=
a
cos
t
,
{\displaystyle x=a\cos t,}
y
=
a
sin
t
{\displaystyle y=a\sin t}
A
(
a
3
2
;
1
2
a
)
{\displaystyle A({a{\sqrt {3}} \over 2};\;{1 \over 2}a)}
B
(
1
2
a
;
a
3
2
)
.
{\displaystyle B({1 \over 2}a;\;{a{\sqrt {3}} \over 2}).}
Įrašę į apskritimo lygtis taškų A ir B koordinates, sužinome, kad tašką A atitinka parametro reikšmė, lygi
π
6
,
{\displaystyle {\pi \over 6},}
B - reikšmė, lygi
π
3
.
{\displaystyle {\pi \over 3}.}
Randame:
d
x
=
−
a
sin
t
d
t
,
{\displaystyle dx=-a\sin tdt,}
d
y
=
a
cos
t
d
t
.
{\displaystyle dy=a\cos tdt.}
∫
L
y
d
x
−
x
d
y
=
∫
π
6
π
3
(
a
sin
t
⋅
(
−
a
)
sin
t
−
a
cos
t
⋅
a
cos
t
)
d
t
=
−
a
2
∫
π
6
π
3
(
cos
2
t
+
sin
2
t
)
d
t
=
−
a
2
∫
π
6
π
3
d
t
=
{\displaystyle \int _{L}ydx-xdy=\int _{\pi \over 6}^{\pi \over 3}(a\sin t\cdot (-a)\sin t-a\cos t\cdot a\cos t)dt=-a^{2}\int _{\pi \over 6}^{\pi \over 3}(\cos ^{2}t+\sin ^{2}t)dt=-a^{2}\int _{\pi \over 6}^{\pi \over 3}dt=}
=
−
a
2
t
|
π
6
π
3
=
−
π
a
2
6
.
{\displaystyle =-a^{2}t|_{\pi \over 6}^{\pi \over 3}=-{\pi a^{2} \over 6}.}
Apskaičiuosime integralą
∫
A
B
x
2
d
x
+
x
y
d
y
,
{\displaystyle \int _{AB}x^{2}dx+xydy,}
AB - ketvirtis apskritimo
x
=
cos
t
,
{\displaystyle x=\cos t,}
y
=
sin
t
,
{\displaystyle y=\sin t,}
0
≤
t
≤
π
2
,
{\displaystyle 0\leq t\leq {\pi \over 2},}
A Atitinka t=0 , B atitinka
t
=
π
/
2.
{\displaystyle t=\pi /2.}
Turime
x
2
=
cos
2
t
,
{\displaystyle x^{2}=\cos ^{2}t,}
d
x
=
−
sin
t
d
t
,
{\displaystyle dx=-\sin tdt,}
x
y
=
cos
t
sin
t
,
{\displaystyle xy=\cos t\sin t,}
d
y
=
cos
t
d
t
.
{\displaystyle dy=\cos tdt.}
∫
A
B
x
2
d
x
+
x
y
d
y
=
∫
0
π
2
(
−
cos
2
t
sin
t
+
cos
2
t
sin
t
)
d
t
=
0.
{\displaystyle \int _{AB}x^{2}dx+xydy=\int _{0}^{\pi \over 2}(-\cos ^{2}t\sin t+\cos ^{2}t\sin t)dt=0.}
Apskaičiuosime integralą
∫
A
B
x
2
d
x
−
y
z
d
y
+
z
d
z
{\displaystyle \int _{AB}x^{2}dx-yzdy+zdz}
AB , jungiančią taškus A(1; 2; -1) ir B(3; 3; 2). Lygtis tiesės AB :
x
−
1
3
−
1
=
y
−
2
3
−
2
=
z
−
(
−
1
)
2
−
(
−
1
)
{\displaystyle {x-1 \over 3-1}={y-2 \over 3-2}={z-(-1) \over 2-(-1)}}
x
−
1
2
=
y
−
2
1
=
z
+
1
3
{\displaystyle {x-1 \over 2}={y-2 \over 1}={z+1 \over 3}}
arba parametrinėje formoje:
x
=
1
+
2
t
,
{\displaystyle x=1+2t,}
y
=
2
+
t
,
{\displaystyle y=2+t,}
z
=
−
1
+
3
t
.
{\displaystyle z=-1+3t.}
AB parametras t keičiasi nuo
t
A
=
0
{\displaystyle t_{A}=0}
t
B
=
1.
{\displaystyle t_{B}=1.}
∫
A
B
x
2
d
x
−
y
z
d
y
+
z
d
z
=
∫
0
1
[
(
1
+
2
t
)
2
⋅
2
−
(
2
+
t
)
(
−
1
+
3
t
)
+
(
−
1
+
3
t
)
3
]
d
t
=
∫
0
1
(
5
t
2
+
12
t
+
1
)
d
t
=
{\displaystyle \int _{AB}x^{2}dx-yzdy+zdz=\int _{0}^{1}[(1+2t)^{2}\cdot 2-(2+t)(-1+3t)+(-1+3t)3]dt=\int _{0}^{1}(5t^{2}+12t+1)dt=}
=
(
5
t
3
3
+
6
t
2
+
t
)
|
0
1
=
26
3
.
{\displaystyle =({5t^{3} \over 3}+6t^{2}+t)|_{0}^{1}={26 \over 3}.}
Duotame pavyzdyje parametru galima parinkti bet kurį iš kintamųjų x , y arba z . Pavyzdžiui, parinkę parametru y , užrašysime lygtį atkarpos AB formoje:
x
=
2
y
−
3
,
{\displaystyle x=2y-3,}
y
=
y
,
{\displaystyle y=y,}
z
=
3
y
−
7
,
{\displaystyle z=3y-7,}
y
A
=
2
≤
y
≤
y
B
=
3.
{\displaystyle y_{A}=2\leq y\leq y_{B}=3.}
Pritaikydami auksčiau išvestą formulę gausime:
∫
A
B
x
2
d
x
−
y
z
d
y
+
z
d
z
=
∫
2
3
[
(
2
y
−
3
)
2
2
−
y
(
3
y
−
7
)
+
(
3
y
−
7
)
3
]
d
y
=
{\displaystyle \int _{AB}x^{2}\;dx-yz\;dy+z\;dz=\int _{2}^{3}[(2y-3)^{2}2-y(3y-7)+(3y-7)3]dy=}
=
∫
2
3
[
2
(
4
y
2
−
12
y
+
9
)
−
3
y
2
+
7
y
+
9
y
−
21
]
d
y
=
∫
2
3
(
5
y
2
−
8
y
−
3
)
d
y
=
{\displaystyle =\int _{2}^{3}[2(4y^{2}-12y+9)-3y^{2}+7y+9y-21]dy=\int _{2}^{3}(5y^{2}-8y-3)dy=}
=
(
5
y
3
3
−
4
y
2
−
3
y
)
|
2
3
=
45
−
36
−
9
−
(
40
3
−
16
−
6
)
=
0
−
(
−
26
3
)
=
26
3
.
{\displaystyle =({5y^{3} \over 3}-4y^{2}-3y)|_{2}^{3}=45-36-9-({40 \over 3}-16-6)=0-(-{26 \over 3})={26 \over 3}.}
Apskaičiuosime integralą
I
=
∫
L
x
y
d
x
+
y
z
d
y
+
z
x
d
z
,
{\displaystyle I=\int _{L}xydx+yzdy+zxdz,}
L - viena apvija spiralinės linijos cilindro paviršiumi
x
=
a
cos
t
,
{\displaystyle x=a\cos t,}
y
=
a
sin
t
,
{\displaystyle y=a\sin t,}
z
=
b
t
{\displaystyle z=bt}
t
0
=
0
{\displaystyle t_{0}=0}
T
=
2
π
.
{\displaystyle T=2\pi .}
Randame:
d
x
=
−
a
sin
t
,
{\displaystyle dx=-a\sin t,}
d
y
=
a
cos
t
,
{\displaystyle dy=a\cos t,}
d
z
=
b
.
{\displaystyle dz=b.}
I
=
∫
0
2
π
(
−
a
3
sin
2
t
cos
t
+
a
2
b
t
sin
t
cos
t
+
a
b
2
t
cos
t
)
d
t
.
{\displaystyle I=\int _{0}^{2\pi }(-a^{3}\sin ^{2}t\cos t+a^{2}bt\sin t\cos t+ab^{2}t\cos t)dt.}
I
1
=
−
a
3
∫
0
2
π
sin
2
t
cos
t
d
t
=
−
a
3
∫
0
2
π
sin
2
t
d
(
sin
t
)
=
−
a
3
sin
3
t
3
|
0
2
π
=
0.
{\displaystyle I_{1}=-a^{3}\int _{0}^{2\pi }\sin ^{2}t\cos t\;dt=-a^{3}\int _{0}^{2\pi }\sin ^{2}t\;d(\sin t)=-a^{3}{\sin ^{3}t \over 3}|_{0}^{2\pi }=0.}
I
2
=
a
2
b
∫
0
2
π
t
sin
t
cos
t
d
t
=
a
2
b
2
∫
0
2
π
t
sin
(
2
t
)
d
t
=
−
a
2
b
4
(
t
cos
(
2
t
)
)
|
0
2
π
+
a
2
b
4
∫
0
2
π
cos
(
2
t
)
d
t
=
{\displaystyle I_{2}=a^{2}b\int _{0}^{2\pi }t\sin t\cos t\;dt={a^{2}b \over 2}\int _{0}^{2\pi }t\sin(2t)dt=-{a^{2}b \over 4}(t\cos(2t))|_{0}^{2\pi }+{a^{2}b \over 4}\int _{0}^{2\pi }\cos(2t)dt=}
=
−
a
2
b
4
(
2
π
cos
(
4
π
)
−
0
cos
0
)
+
a
2
b
8
sin
(
2
t
)
|
0
2
π
=
−
π
a
2
b
2
,
{\displaystyle =-{a^{2}b \over 4}(2\pi \cos(4\pi )-0\cos 0)+{a^{2}b \over 8}\sin(2t)|_{0}^{2\pi }=-{\pi a^{2}b \over 2},}
u
=
t
,
{\displaystyle u=t,}
d
v
=
sin
(
2
t
)
,
{\displaystyle dv=\sin(2t),}
d
u
=
d
t
,
{\displaystyle du=dt,}
v
=
−
cos
(
2
t
)
2
.
{\displaystyle v=-{\cos(2t) \over 2}.}
I
3
=
a
b
2
∫
0
2
π
t
cos
t
d
t
=
a
b
2
(
t
sin
t
)
|
0
2
π
−
a
b
2
∫
0
2
π
sin
t
d
t
=
a
b
2
cos
t
|
0
2
π
=
0.
{\displaystyle I_{3}=ab^{2}\int _{0}^{2\pi }t\cos t\;dt=ab^{2}(t\sin t)|_{0}^{2\pi }-ab^{2}\int _{0}^{2\pi }\sin tdt=ab^{2}\cos t|_{0}^{2\pi }=0.}
u
=
t
,
{\displaystyle u=t,}
d
v
=
cos
t
,
{\displaystyle dv=\cos t,}
d
u
=
d
t
,
{\displaystyle du=dt,}
v
=
sin
t
.
{\displaystyle v=\sin t.}
Todėl
I
=
I
1
+
I
2
+
I
3
=
0
−
π
a
2
b
2
+
0
=
−
π
a
2
b
2
.
{\displaystyle I=I_{1}+I_{2}+I_{3}=0-{\pi a^{2}b \over 2}+0=-{\pi a^{2}b \over 2}.}
Apskaičiuoti darbą jėgos
F
→
(
x
;
y
)
,
{\displaystyle {\vec {F}}(x;\;y),}
Sprendimas . Pagal sąlygą,
|
F
→
(
x
;
y
)
|
=
x
2
+
y
2
;
{\displaystyle |{\vec {F}}(x;y)|={\sqrt {x^{2}+y^{2}}};}
F
→
(
x
;
y
)
{\displaystyle {\vec {F}}(x;\;y)}
P
=
−
x
,
Q
=
−
y
{\displaystyle P=-x,\;\;Q=-y}
−
{\displaystyle -}
A
=
∫
B
C
P
d
x
+
Q
d
y
{\displaystyle A=\int _{BC}Pdx+Qdy}
A
=
−
∮
L
x
d
x
+
y
d
y
,
{\displaystyle A=-\oint _{L}x\;dx+y\;dy,}
L - elipsė
x
=
a
cos
(
t
)
,
{\displaystyle x=a\cos(t),}
y
=
b
sin
(
t
)
,
{\displaystyle y=b\sin(t),}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
d
x
=
−
a
sin
(
t
)
d
t
,
{\displaystyle dx=-a\sin(t)dt,}
d
y
=
b
cos
(
t
)
d
t
{\displaystyle dy=b\cos(t)dt}
A
=
−
∮
L
x
d
x
+
y
d
y
=
−
∫
0
2
π
x
d
x
−
∫
0
2
π
y
d
y
=
−
∫
0
2
π
−
a
2
cos
(
t
)
sin
(
t
)
d
t
−
∫
0
2
π
b
sin
(
t
)
b
cos
(
t
)
d
t
=
{\displaystyle A=-\oint _{L}x\;dx+y\;dy=-\int _{0}^{2\pi }x\;dx-\int _{0}^{2\pi }y\;dy=-\int _{0}^{2\pi }-a^{2}\cos(t)\sin(t)dt-\int _{0}^{2\pi }b\sin(t)b\cos(t)dt=}
=
∫
0
2
π
a
2
cos
(
t
)
sin
(
t
)
d
t
−
∫
0
2
π
b
2
sin
(
t
)
cos
(
t
)
d
t
=
∫
0
2
π
(
a
2
cos
(
t
)
sin
(
t
)
−
b
2
sin
(
t
)
cos
(
t
)
)
d
t
=
{\displaystyle =\int _{0}^{2\pi }a^{2}\cos(t)\sin(t)dt-\int _{0}^{2\pi }b^{2}\sin(t)\cos(t)dt=\int _{0}^{2\pi }(a^{2}\cos(t)\sin(t)-b^{2}\sin(t)\cos(t))dt=}
=
(
a
2
−
b
2
)
∫
0
2
π
sin
(
t
)
cos
(
t
)
d
t
=
a
2
−
b
2
2
∫
0
2
π
sin
(
2
t
)
d
t
=
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
2
π
=
{\displaystyle =(a^{2}-b^{2})\int _{0}^{2\pi }\sin(t)\cos(t)dt={\frac {a^{2}-b^{2}}{2}}\int _{0}^{2\pi }\sin(2t)dt={\frac {a^{2}-b^{2}}{4}}(-\cos(2t))|_{0}^{2\pi }=}
=
−
a
2
−
b
2
4
(
cos
(
4
π
)
−
cos
(
2
⋅
0
)
)
=
−
a
2
−
b
2
4
(
1
−
1
)
=
0.
{\displaystyle =-{\frac {a^{2}-b^{2}}{4}}(\cos(4\pi )-\cos(2\cdot 0))=-{\frac {a^{2}-b^{2}}{4}}(1-1)=0.}
Pastebėsime, kad iš to, kad integralas pasirodė lygus nuliui, seka, kad pointegralinė išraiška yra pilnas diferencialas tam tikros funkcijos (raskite šią funkciją savarankiškai). [ keisti ] Jei kreivė AB išreikšta lygtimi
y
=
y
(
x
)
{\displaystyle y=y(x)}
a
≤
x
≤
b
,
{\displaystyle a\leq x\leq b,}
y
(
x
)
{\displaystyle y(x)}
∫
A
B
P
(
x
,
y
)
d
x
+
Q
(
x
,
y
)
d
y
=
∫
a
b
[
P
[
x
,
y
(
x
)
]
+
Q
[
x
,
y
(
x
)
]
y
′
(
x
)
]
d
x
.
{\displaystyle \int _{AB}P(x,y)dx+Q(x,y)dy=\int _{a}^{b}[P[x,y(x)]+Q[x,y(x)]y'(x)]dx.}
Apskaičiuosime integralą:
∫
A
B
x
y
d
x
−
(
x
−
y
)
d
y
{\displaystyle \int _{AB}xydx-(x-y)dy}
AB prabolės
y
=
x
2
{\displaystyle y=x^{2}}
x
A
=
−
1
,
{\displaystyle x_{A}=-1,}
x
B
=
−
2.
{\displaystyle x_{B}=-2.}
x ir pakeitę
d
y
=
2
x
d
x
,
{\displaystyle dy=2xdx,}
∫
A
B
x
y
d
x
−
(
x
−
y
)
d
y
=
∫
−
1
−
2
[
x
⋅
x
2
−
(
x
−
x
2
)
⋅
2
x
]
d
x
=
∫
−
1
−
2
(
3
x
3
−
2
x
2
)
d
x
=
(
3
4
x
4
−
2
3
x
3
)
|
−
1
−
2
=
{\displaystyle \int _{AB}xy\;dx-(x-y)dy=\int _{-1}^{-2}[x\cdot x^{2}-(x-x^{2})\cdot 2x]dx=\int _{-1}^{-2}(3x^{3}-2x^{2})dx=({3 \over 4}x^{4}-{2 \over 3}x^{3})|_{-1}^{-2}=}
=
12
+
16
3
−
(
3
4
+
2
3
)
=
45
4
+
14
3
=
191
12
=
15
11
12
.
{\displaystyle =12+{16 \over 3}-({3 \over 4}+{2 \over 3})={45 \over 4}+{14 \over 3}={191 \over 12}=15{11 \over 12}.}
Apskaičiuosime integralą
I
=
∫
L
y
2
d
x
+
(
x
y
−
x
2
)
d
y
,
{\displaystyle I=\int _{L}y^{2}dx+(xy-x^{2})dy,}
L - lankas prabolės
y
2
=
9
x
{\displaystyle y^{2}=9x}
A(0; 0) iki taško B(1; 3) .
x
=
y
2
9
,
{\displaystyle x={y^{2} \over 9},}
d
x
=
2
y
9
.
{\displaystyle dx={2y \over 9}.}
I
=
∫
0
3
[
2
9
y
3
+
(
y
3
9
−
y
4
81
)
]
d
y
=
∫
0
3
(
y
3
3
−
y
4
81
)
d
y
=
(
y
4
12
−
y
5
405
)
|
0
3
=
27
4
−
243
405
=
9963
1620
=
6
3
20
=
6.15.
{\displaystyle I=\int _{0}^{3}[{2 \over 9}y^{3}+({y^{3} \over 9}-{y^{4} \over 81})]dy=\int _{0}^{3}({y^{3} \over 3}-{y^{4} \over 81})dy=({y^{4} \over 12}-{y^{5} \over 405})|_{0}^{3}={27 \over 4}-{243 \over 405}={9963 \over 1620}=6{3 \over 20}=6.15.}
Kvadratas. Apskaičiuosime integralą
∮
L
(
x
+
y
)
d
y
,
{\displaystyle \oint _{L}(x+y)dy,}
L - konturas stačiakampio, padaryto iš tiesių
x
=
0
,
{\displaystyle x=0,}
y
=
0
,
{\displaystyle y=0,}
x
=
1
{\displaystyle x=1}
y
=
1.
{\displaystyle y=1.}
Paveiksle teigiama kryptis apėjimo konturo L paženklinta rodyklėmis. Padalinę visą kontūrą integravimo į dalis, užrašysime:
∮
L
=
∫
A
B
+
∫
B
C
+
∫
C
D
+
∫
D
A
.
{\displaystyle \oint _{L}=\int _{AB}+\int _{BC}+\int _{CD}+\int _{DA}.}
AB ir CD lygus nuliui, todėl, kad ant jų y yra pastovus ir, dėl to
d
y
=
0.
{\displaystyle dy=0.}
BC ir DA . Pagal formulę [ vietoje x įrašę y ir vietoje y(x) įrašę x(y)], gauname
∫
B
C
(
x
+
y
)
d
y
=
∫
0
1
(
1
+
y
)
d
y
=
(
y
+
y
2
2
)
|
0
1
=
3
2
.
{\displaystyle \int _{BC}(x+y)dy=\int _{0}^{1}(1+y)dy=(y+{y^{2} \over 2})|_{0}^{1}={3 \over 2}.}
∫
D
A
(
x
+
y
)
d
y
=
∫
1
0
(
0
+
y
)
d
y
=
y
2
2
|
1
0
=
−
1
2
.
{\displaystyle \int _{DA}(x+y)dy=\int _{1}^{0}(0+y)dy={y^{2} \over 2}|_{1}^{0}=-{1 \over 2}.}
∮
(
x
+
y
)
d
y
=
3
2
−
1
2
=
1.
{\displaystyle \oint (x+y)dy={3 \over 2}-{1 \over 2}=1.}
Trikampis trimatėje erdvėje. Apskaičiuosime integralą
∮
(
x
+
y
+
z
)
d
x
{\displaystyle \oint (x+y+z)dx}
ABCA su viršūnėmis A (1; 0; 0), B (0; 1; 0), C (0; 0; 1). Pagal apibrėžimą
∮
A
B
C
A
=
∫
A
B
+
∫
B
C
+
∫
C
A
.
{\displaystyle \oint _{ABCA}=\int _{AB}+\int _{BC}+\int _{CA}.}
∫
B
C
(
x
+
y
+
z
)
=
0
,
{\displaystyle \int _{BC}(x+y+z)=0,}
yOz , statmenoje ašiai Ox (todėl
x
=
0
{\displaystyle x=0}
d
x
=
0
)
.
{\displaystyle dx=0).}
AB užrašysime pavidale:
x
=
x
,
{\displaystyle x=x,}
y
=
1
−
x
,
{\displaystyle y=1-x,}
z
=
0.
{\displaystyle z=0.}
x
A
=
1
,
{\displaystyle x_{A}=1,}
x
B
=
0
,
{\displaystyle x_{B}=0,}
∫
A
B
=
(
x
+
y
+
z
)
d
x
=
∫
1
0
[
x
+
(
1
−
x
)
+
0
]
d
x
=
∫
1
0
d
x
=
x
|
1
0
=
−
1.
{\displaystyle \int _{AB}=(x+y+z)dx=\int _{1}^{0}[x+(1-x)+0]dx=\int _{1}^{0}dx=x|_{1}^{0}=-1.}
Lygtį atkarpos CA užrašysime pavidale:
x
=
x
,
{\displaystyle x=x,}
y
=
0
,
{\displaystyle y=0,}
z
=
1
−
x
.
{\displaystyle z=1-x.}
x
C
=
0
,
{\displaystyle x_{C}=0,}
x
A
=
1
,
{\displaystyle x_{A}=1,}
∫
C
A
(
x
+
y
+
z
)
d
x
=
∫
0
1
[
x
+
(
1
−
x
)
]
d
x
=
1.
{\displaystyle \int _{CA}(x+y+z)dx=\int _{0}^{1}[x+(1-x)]dx=1.}
∮
A
B
C
A
(
x
+
y
+
z
)
d
x
=
−
1
+
0
+
1
=
0.
{\displaystyle \oint {ABCA}(x+y+z)dx=-1+0+1=0.}
Parabolė. Apskaičiuosime integralą
∫
A
O
B
y
d
x
,
{\displaystyle \int _{AOB}ydx,}
AOB - lankas prabolės
y
2
=
x
,
{\displaystyle y^{2}=x,}
A (1; -1), B (1; 1). Atsižvelgiant į savybes kreivinių integralų, gausime:
∫
A
O
B
y
d
x
=
∫
A
O
y
d
x
+
∫
O
B
y
d
x
.
{\displaystyle \int _{AOB}ydx=\int _{AO}ydx+\int _{OB}ydx.}
AO lygtis yra
y
=
−
x
,
{\displaystyle y=-{\sqrt {x}},}
0
≤
x
≤
1
,
{\displaystyle 0\leq x\leq 1,}
x
A
=
1
,
{\displaystyle x_{A}=1,}
x
O
=
0
,
{\displaystyle x_{O}=0,}
OB turi lygtį
y
=
x
,
{\displaystyle y={\sqrt {x}},}
0
≤
x
≤
1
,
{\displaystyle 0\leq x\leq 1,}
x
0
=
0
,
{\displaystyle x_{0}=0,}
x
B
=
1
,
{\displaystyle x_{B}=1,}
∫
A
O
B
y
d
x
=
∫
1
0
−
x
d
x
+
∫
0
1
x
d
x
=
2
∫
0
1
x
d
x
=
4
3
x
3
2
|
0
1
=
4
3
.
{\displaystyle \int _{AOB}ydx=\int _{1}^{0}-{\sqrt {x}}dx+\int _{0}^{1}{\sqrt {x}}dx=2\int _{0}^{1}{\sqrt {x}}dx={4 \over 3}x^{3 \over 2}|_{0}^{1}={4 \over 3}.}
Šiame pavyzdyje paprasčiau naudoti parametrą y , pakeičiant atitinkamai formulę:
∫
A
O
B
y
d
x
=
∫
−
1
1
y
⋅
2
y
d
y
=
2
3
y
3
|
−
1
1
=
4
3
.
{\displaystyle \int _{AOB}y\;dx=\int _{-1}^{1}y\cdot 2y\;dy={2 \over 3}y^{3}|_{-1}^{1}={4 \over 3}.}
Integravimo keliai sutampa. Apskaičiuosime integralą
∫
A
B
3
x
2
y
d
x
+
(
x
3
+
1
)
d
y
,
{\displaystyle \int _{AB}3x^{2}y\;dx+(x^{3}+1)dy,}
a) AB - tiesė
y
=
x
,
{\displaystyle y=x,}
b) AB - parabolė y=x², sujungianti tuos pačius taškus (0; 0) ir (1; 1);
c) AB - laužtė, pereinanti per taškus (0; 0), (1; 0), (1; 1).
Pagal vieno iš kinamųjų pakeitimo formulę turime:
a)
∫
A
B
3
x
2
y
d
x
+
(
x
3
+
1
)
d
y
=
∫
0
1
[
3
x
2
⋅
x
+
(
x
3
+
1
)
]
d
x
=
{\displaystyle \int _{AB}3x^{2}y\;dx+(x^{3}+1)dy=\int _{0}^{1}[3x^{2}\cdot x+(x^{3}+1)]dx=}
=
∫
0
1
(
4
x
3
+
1
)
d
x
=
(
x
4
+
x
)
|
0
1
=
2
;
{\displaystyle =\int _{0}^{1}(4x^{3}+1)dx=(x^{4}+x)|_{0}^{1}=2;}
b)
∫
A
B
3
x
2
y
d
x
+
(
x
3
+
1
)
d
y
=
∫
0
1
[
3
x
2
⋅
x
2
+
(
x
3
+
1
)
2
x
]
d
x
=
∫
0
1
(
5
x
4
+
2
x
)
d
x
=
(
x
5
+
x
2
)
|
0
1
=
2
;
{\displaystyle \int _{AB}3x^{2}y\;dx+(x^{3}+1)dy=\int _{0}^{1}[3x^{2}\cdot x^{2}+(x^{3}+1)2x]dx=\int _{0}^{1}(5x^{4}+2x)dx=(x^{5}+x^{2})|_{0}^{1}=2;}
c)
∫
A
B
3
x
2
y
d
x
+
(
x
3
+
1
)
d
y
=
∫
0
1
3
x
2
⋅
0
d
x
+
∫
0
1
(
1
+
1
)
d
y
=
∫
0
1
2
d
y
=
2.
{\displaystyle \int _{AB}3x^{2}y\;dx+(x^{3}+1)dy=\int _{0}^{1}3x^{2}\cdot 0\;dx+\int _{0}^{1}(1+1)dy=\int _{0}^{1}2dy=2.}
Vienodo atsakymo gavimas integruojant skirtingais keliais yra dėl to, kad
∂
P
∂
y
=
∂
Q
∂
x
,
{\displaystyle {\partial P \over \partial y}={\partial Q \over \partial x},}
∂
(
3
x
2
y
)
∂
y
=
3
x
2
=
∂
(
x
3
+
1
)
∂
x
.
{\displaystyle {\partial (3x^{2}y) \over \partial y}=3x^{2}={\partial (x^{3}+1) \over \partial x}.}
Apskaičiuokime integralą
∫
L
x
y
d
x
+
(
x
2
−
y
3
)
d
y
{\displaystyle \int _{L}xydx+(x^{2}-y^{3})dy}
O (0; 0) iki taško A(1; 1), kai integravimo kelias L nusakomas lygtimi: a)
y
=
x
;
{\displaystyle y=x;}
y
=
x
3
;
{\displaystyle y=x^{3};}
y
2
=
x
.
{\displaystyle y^{2}=x.}
a) Randame
d
y
=
d
x
.
{\displaystyle dy=dx.}
∫
L
x
y
d
x
+
(
x
2
−
y
3
)
d
y
=
∫
0
1
x
⋅
x
d
x
+
∫
0
1
(
x
2
−
x
3
)
d
x
=
{\displaystyle \int _{L}xydx+(x^{2}-y^{3})dy=\int _{0}^{1}x\cdot x\;dx+\int _{0}^{1}(x^{2}-x^{3})dx=}
=
∫
0
1
(
2
x
2
−
x
3
)
d
x
=
(
2
x
3
3
−
x
4
4
)
|
0
1
=
5
12
.
{\displaystyle =\int _{0}^{1}(2x^{2}-x^{3})dx=({2x^{3} \over 3}-{x^{4} \over 4})|_{0}^{1}={5 \over 12}.}
b) Kadangi
d
y
=
3
x
2
d
x
,
{\displaystyle dy=3x^{2}dx,}
∫
L
x
y
d
x
+
(
x
2
−
y
3
)
d
y
=
∫
0
1
x
⋅
x
3
d
x
+
∫
0
1
(
x
2
−
x
9
)
⋅
3
x
2
d
x
=
∫
0
1
(
4
x
4
−
3
x
11
)
d
x
=
(
4
5
x
5
−
x
12
4
)
|
0
1
=
11
20
.
{\displaystyle \int _{L}xydx+(x^{2}-y^{3})dy=\int _{0}^{1}x\cdot x^{3}dx+\int _{0}^{1}(x^{2}-x^{9})\cdot 3x^{2}dx=\int _{0}^{1}(4x^{4}-3x^{11})dx=({4 \over 5}x^{5}-{x^{12} \over 4})|_{0}^{1}={11 \over 20}.}
c) Iš sąlygos
y
2
=
x
{\displaystyle y^{2}=x}
y
=
x
,
{\displaystyle y={\sqrt {x}},}
d
y
=
1
2
x
−
1
2
d
x
.
{\displaystyle dy={1 \over 2}x^{-{1 \over 2}}dx.}
∫
L
x
y
d
x
+
(
x
2
−
y
3
)
d
y
=
∫
0
1
x
x
d
x
+
∫
0
1
(
x
2
−
x
x
)
d
x
2
x
=
∫
0
1
(
3
2
x
3
2
−
x
2
)
d
x
=
(
3
5
x
5
2
−
1
4
x
2
)
|
0
1
=
7
20
.
{\displaystyle \int _{L}xydx+(x^{2}-y^{3})dy=\int _{0}^{1}x{\sqrt {x}}dx+\int _{0}^{1}(x^{2}-x{\sqrt {x}}){dx \over 2{\sqrt {x}}}=\int _{0}^{1}({3 \over 2}x^{3 \over 2}-{x \over 2})dx=({3 \over 5}x^{5 \over 2}-{1 \over 4}x^{2})|_{0}^{1}={7 \over 20}.}
Šiuo atveju pavyzdį galėjome spręsti ir neišreikšdami kintąmąjį y kintamuoju x . Laikykime x funkcija, o y argumentu. Tuomet iš sąlygos
x
=
y
2
{\displaystyle x=y^{2}}
d
x
=
2
y
d
y
;
{\displaystyle dx=2ydy;}
y kitimo rėžiai yra nuo 0 iki 1. Dabar duotąjį integralą pakeiskime apibrėžtiniu, įrašydami vietoje x ir dx jų išraiškas:
∫
L
x
y
d
x
+
(
x
2
−
y
3
)
d
y
=
∫
0
1
y
2
⋅
y
⋅
2
y
d
y
+
∫
0
1
(
y
4
−
y
3
)
d
y
=
∫
0
1
(
3
y
4
−
y
3
)
d
y
=
(
3
5
y
5
−
y
4
4
)
|
0
1
=
7
20
.
{\displaystyle \int _{L}xydx+(x^{2}-y^{3})dy=\int _{0}^{1}y^{2}\cdot y\cdot 2ydy+\int _{0}^{1}(y^{4}-y^{3})dy=\int _{0}^{1}(3y^{4}-y^{3})dy=({3 \over 5}y^{5}-{y^{4} \over 4})|_{0}^{1}={7 \over 20}.}
Visais triais atvejais integravimo pradžia ir pabaiga sutapo, tačiau integruodami galvome skirtingus atsakymus, nes
∂
P
∂
y
≠
∂
Q
∂
x
,
{\displaystyle {\partial P \over \partial y}\neq {\partial Q \over \partial x},}
∂
(
x
y
)
∂
y
=
x
≠
2
x
=
∂
(
x
2
−
y
3
)
∂
x
.
{\displaystyle {\partial (xy) \over \partial y}=x\neq 2x={\partial (x^{2}-y^{3}) \over \partial x}.}
Prabolė, tiesė AB ir laužtė ACB .
∫
L
(
2
x
+
3
y
2
)
d
x
+
(
6
x
y
−
1
)
d
y
{\displaystyle \int _{L}(2x+3y^{2})dx+(6xy-1)dy}
A (1; 1) iki taško B (2; 4), kai integravimo kelias L nusakomas:
a) tiesės
y
=
3
x
−
2
{\displaystyle y=3x-2}
b) parabolės
y
=
x
2
{\displaystyle y=x^{2}}
c) laužte ACB . a) Iš lygties
y
=
3
x
−
2
{\displaystyle y=3x-2}
d
y
=
3
d
x
;
{\displaystyle dy=3dx;}
∫
L
(
2
x
+
3
y
2
)
d
x
+
(
6
x
y
−
1
)
d
y
=
∫
1
2
(
2
x
+
27
x
2
−
36
x
+
12
)
d
x
+
∫
1
2
(
18
x
2
−
12
x
−
1
)
⋅
3
x
d
x
=
{\displaystyle \int _{L}(2x+3y^{2})dx+(6xy-1)dy=\int _{1}^{2}(2x+27x^{2}-36x+12)dx+\int _{1}^{2}(18x^{2}-12x-1)\cdot 3xdx=}
=
∫
1
2
(
81
x
2
−
70
x
+
9
)
d
x
=
(
27
x
3
−
35
x
2
+
9
x
)
|
1
2
=
93.
{\displaystyle =\int _{1}^{2}(81x^{2}-70x+9)dx=(27x^{3}-35x^{2}+9x)|_{1}^{2}=93.}
b) kai
y
=
x
2
,
{\displaystyle y=x^{2},}
d
y
=
2
x
d
x
{\displaystyle dy=2xdx}
∫
L
(
2
x
+
3
y
2
)
d
x
+
(
6
x
y
−
1
)
d
y
=
∫
1
2
(
2
x
+
3
x
4
)
d
x
+
∫
1
2
(
6
x
3
−
1
)
2
x
d
x
=
∫
1
2
15
x
4
d
x
=
3
x
5
|
1
2
=
93.
{\displaystyle \int _{L}(2x+3y^{2})dx+(6xy-1)dy=\int _{1}^{2}(2x+3x^{4})dx+\int _{1}^{2}(6x^{3}-1)2xdx=\int _{1}^{2}15x^{4}dx=3x^{5}|_{1}^{2}=93.}
c) Integravimo kelią suskaidysime į dvi atkarpas: AC ir CB . Atkarpos AC taškuose x kinta nuo 1 iki 2,
y
=
1
=
c
o
n
s
t
,
{\displaystyle y=1=const,}
AC taškuose
d
y
=
0
;
{\displaystyle dy=0;}
CB taškuose y kinta nuo 1 iki 4,
x
=
2
=
c
o
n
s
t
,
{\displaystyle x=2=const,}
CB taškuose
d
x
=
0.
{\displaystyle dx=0.}
∫
L
(
2
x
+
3
y
2
)
d
x
+
(
6
x
y
−
1
)
d
y
=
∫
A
C
(
2
x
+
3
y
2
)
d
x
+
(
6
x
y
−
1
)
d
y
+
∫
C
B
(
2
x
+
3
y
2
)
d
x
+
(
6
x
y
−
1
)
d
y
=
{\displaystyle \int _{L}(2x+3y^{2})dx+(6xy-1)dy=\int _{AC}(2x+3y^{2})dx+(6xy-1)dy+\int _{CB}(2x+3y^{2})dx+(6xy-1)dy=}
=
∫
1
2
(
2
x
+
3
⋅
1
2
)
d
x
+
(
6
x
⋅
1
−
1
)
⋅
0
+
∫
1
4
(
2
⋅
2
+
3
y
2
)
⋅
0
+
(
6
⋅
2
−
1
)
d
y
=
∫
1
2
(
2
x
+
3
)
d
x
+
∫
1
4
(
12
y
−
1
)
d
y
=
{\displaystyle =\int _{1}^{2}(2x+3\cdot 1^{2})dx+(6x\cdot 1-1)\cdot 0+\int _{1}^{4}(2\cdot 2+3y^{2})\cdot 0+(6\cdot 2-1)dy=\int _{1}^{2}(2x+3)dx+\int _{1}^{4}(12y-1)dy=}
=
(
x
2
+
3
)
|
1
2
+
(
6
y
2
−
y
)
|
1
4
=
6
+
87
=
93.
{\displaystyle =(x^{2}+3)|_{1}^{2}+(6y^{2}-y)|_{1}^{4}=6+87=93.}
Visais variantais gavome tokį patį kreivinio integralo atsakymą. Sakoma, kad krivinis integralas nepriklauso nuo integravimo kelio, o priklauso tik nuo integravimo kreivės pradžios ir pabaigos taškų. Taip yra todėl, kad
∂
P
∂
y
=
∂
(
2
x
+
3
y
2
)
∂
y
=
6
y
=
∂
(
6
x
y
−
1
)
∂
x
=
∂
Q
∂
x
.
{\displaystyle {\partial P \over \partial y}={\partial (2x+3y^{2}) \over \partial y}=6y={\partial (6xy-1) \over \partial x}={\partial Q \over \partial x}.}
[ keisti ] Kad integralas
∫
L
P
(
x
,
y
)
d
x
+
Q
(
x
,
y
)
d
y
{\displaystyle \int _{L}P(x,y)dx+Q(x,y)dy}
∂
P
(
x
,
y
)
∂
y
=
∂
Q
(
x
,
y
)
∂
x
{\displaystyle {\partial P(x,y) \over \partial y}={\partial Q(x,y) \over \partial x}}
∂
P
∂
y
−
∂
Q
∂
x
=
0.
{\displaystyle {\partial P \over \partial y}-{\partial Q \over \partial x}=0.}
Trimatis vektorius su dedamosiomis (projekcijomis koordinačių ašims)
P
(
x
,
y
,
z
)
,
{\displaystyle P(x,y,z),}
Q
(
x
,
y
,
z
)
,
{\displaystyle Q(x,y,z),}
R
(
x
,
y
,
z
)
{\displaystyle R(x,y,z)}
∫
A
B
P
d
x
+
Q
d
y
+
R
d
z
,
{\displaystyle \int _{AB}Pdx+Qdy+Rdz,}
∂
P
∂
y
=
∂
Q
∂
x
,
∂
P
∂
z
=
∂
R
∂
x
,
∂
Q
∂
z
=
∂
R
∂
y
.
{\displaystyle {\partial P \over \partial y}={\partial Q \over \partial x},\;{\partial P \over \partial z}={\partial R \over \partial x},\;{\partial Q \over \partial z}={\partial R \over \partial y}.}
Prabolė ir tiesė. Apskaičiuosime, pavyzdžiui, integralą
∫
A
B
x
d
x
+
x
y
d
y
+
y
d
z
{\displaystyle \int _{AB}xdx+xydy+ydz}
AB , jungiančios taškus A (1; 0; 0) ir B (1; 1; 1). Lygtis tiesės AB :
x
−
1
1
−
1
=
y
−
0
1
−
0
=
z
−
0
1
−
0
,
{\displaystyle {x-1 \over 1-1}={y-0 \over 1-0}={z-0 \over 1-0},}
x
−
1
0
=
y
−
0
1
=
z
−
0
1
,
{\displaystyle {x-1 \over 0}={y-0 \over 1}={z-0 \over 1},}
t. y.
y
=
z
,
{\displaystyle y=z,}
x
−
1
=
0
,
{\displaystyle x-1=0,}
x
=
1.
{\displaystyle x=1.}
y , turime:
∫
A
B
x
d
x
+
x
y
d
y
+
y
d
z
=
∫
0
1
1
⋅
0
+
1
⋅
y
d
y
+
y
d
y
=
∫
0
1
2
y
d
y
=
y
2
|
0
1
=
1.
{\displaystyle \int _{AB}xdx+xydy+ydz=\int _{0}^{1}1\cdot 0+1\cdot ydy+ydy=\int _{0}^{1}2ydy=y^{2}|_{0}^{1}=1.}
Apskaičiuosime dabar tą patį integralą pagal lanką prabolės AB , aprašamos lygtimis
x
=
1
,
{\displaystyle x=1,}
z
=
y
2
.
{\displaystyle z=y^{2}.}
y (
x
=
1
,
{\displaystyle x=1,}
y
=
y
,
{\displaystyle y=y,}
z
=
y
2
,
{\displaystyle z=y^{2},}
d
z
=
2
y
d
y
,
{\displaystyle dz=2ydy,}
y
A
=
0
,
{\displaystyle y_{A}=0,}
y
B
=
1
{\displaystyle y_{B}=1}
∫
A
B
x
d
x
+
x
y
d
y
+
y
d
z
=
∫
0
1
1
⋅
0
+
1
⋅
y
d
y
+
y
⋅
2
y
=
∫
0
1
(
y
+
2
y
2
)
d
y
=
(
y
2
2
+
2
y
3
3
)
|
0
1
=
1
⋅
3
+
2
⋅
2
2
⋅
3
=
7
6
.
{\displaystyle \int _{AB}xdx+xydy+ydz=\int _{0}^{1}1\cdot 0+1\cdot ydy+y\cdot 2y=\int _{0}^{1}(y+2y^{2})dy=({y^{2} \over 2}+{2y^{3} \over 3})|_{0}^{1}={1\cdot 3+2\cdot 2 \over 2\cdot 3}={7 \over 6}.}
Šis pavyzdys parodo, kad integralo
∫
A
B
x
d
x
+
x
y
d
y
+
y
d
z
{\displaystyle \int _{AB}xdx+xydy+ydz}
∂
P
∂
y
=
0
≠
y
=
∂
Q
∂
x
,
∂
P
∂
z
=
0
=
∂
R
∂
x
,
∂
Q
∂
z
=
0
≠
1
=
∂
R
∂
y
.
{\displaystyle {\partial P \over \partial y}=0\neq y={\partial Q \over \partial x},\;{\partial P \over \partial z}=0={\partial R \over \partial x},\;{\partial Q \over \partial z}=0\neq 1={\partial R \over \partial y}.}
![{\displaystyle =\int _{\alpha }^{\beta }[P(x(t),y(t),z(t))x'(t)+Q(x(t),y(t),z(t))y'(t)+R(x(t),y(t),z(t))z'(t)]dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4266861ffc788abc70a956b63d656cc1a77d7da)
![{\displaystyle \int _{AB}x^{2}dx-yzdy+zdz=\int _{0}^{1}[(1+2t)^{2}\cdot 2-(2+t)(-1+3t)+(-1+3t)3]dt=\int _{0}^{1}(5t^{2}+12t+1)dt=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08d1ac8a2dba684fb5074d21f1f3a142b06df411)
![{\displaystyle \int _{AB}x^{2}\;dx-yz\;dy+z\;dz=\int _{2}^{3}[(2y-3)^{2}2-y(3y-7)+(3y-7)3]dy=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0fc164f522ddbfce48dee7511d6fac42b3350f94)
![{\displaystyle =\int _{2}^{3}[2(4y^{2}-12y+9)-3y^{2}+7y+9y-21]dy=\int _{2}^{3}(5y^{2}-8y-3)dy=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b3bde4f582c41d63029124e539f0768fedd7d85)
![{\displaystyle \int _{AB}P(x,y)dx+Q(x,y)dy=\int _{a}^{b}[P[x,y(x)]+Q[x,y(x)]y'(x)]dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f278004217b0994e88fdfb6515243b003752dc06)
![{\displaystyle \int _{AB}xy\;dx-(x-y)dy=\int _{-1}^{-2}[x\cdot x^{2}-(x-x^{2})\cdot 2x]dx=\int _{-1}^{-2}(3x^{3}-2x^{2})dx=({3 \over 4}x^{4}-{2 \over 3}x^{3})|_{-1}^{-2}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1feb0665ce2db40674b6795fc3cad1ce094b9478)
![{\displaystyle I=\int _{0}^{3}[{2 \over 9}y^{3}+({y^{3} \over 9}-{y^{4} \over 81})]dy=\int _{0}^{3}({y^{3} \over 3}-{y^{4} \over 81})dy=({y^{4} \over 12}-{y^{5} \over 405})|_{0}^{3}={27 \over 4}-{243 \over 405}={9963 \over 1620}=6{3 \over 20}=6.15.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50f8bee9033ad13fb94d22f3ca1fb20c0ec8db1a)
![{\displaystyle \int _{AB}=(x+y+z)dx=\int _{1}^{0}[x+(1-x)+0]dx=\int _{1}^{0}dx=x|_{1}^{0}=-1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e4611f1963d716e00a86c8152106429426ab743)
![{\displaystyle \int _{CA}(x+y+z)dx=\int _{0}^{1}[x+(1-x)]dx=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a7fe0338df9dc04fdcbc92c17350114e8cb57b9)
![{\displaystyle \int _{AB}3x^{2}y\;dx+(x^{3}+1)dy=\int _{0}^{1}[3x^{2}\cdot x+(x^{3}+1)]dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb40e04e68fbe1d9e138a18e646447a019273fa9)
![{\displaystyle \int _{AB}3x^{2}y\;dx+(x^{3}+1)dy=\int _{0}^{1}[3x^{2}\cdot x^{2}+(x^{3}+1)2x]dx=\int _{0}^{1}(5x^{4}+2x)dx=(x^{5}+x^{2})|_{0}^{1}=2;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b607c3c90cfbded5d3d12f65166d832c824d950)
