Maple dasturi yordamida aniqmas va aniq integrallarni hisoblash

int(x*y-x,x=0..1);

1 _{y } 1
2 2

• int(sin(x*y),x=1..Pi);

• 
  int(x+y/x,y=-1..1);
  

• 
  cos(  y )  cos( y )
  

y

2 x

Yuqorida keltirilgan aniq integral parametrga bog’liq aniq integrallarning xususiy holi hisoblanadi. Umumiy holda parametrga bog’liq integrallar quyidagicha ta’riflanadi:
Agar 𝑢⁽𝑦⁾ va 𝑤⁽𝑦⁾ funksiyalar biror G to’plamda aniqlangan bo’lib,
𝑤⁽𝑦⁾
𝐹⁽𝑦⁾ = ∫ 𝑓⁽𝑥, 𝑦⁾𝑑𝑥
𝑢⁽𝑦⁾
integral mavjud bo’lsa, 𝐹⁽𝑦⁾ funksiyaga 𝒚 parametrga bog’liq aniq integral deyiladi. 𝐹⁽𝑦⁾
funkiyaning hosilasi quyidagi formuladan topiladi:
𝑤⁽𝑦^{)
𝜕𝐹(𝑦)} = ∫ ^{𝜕(𝑓(𝑥, 𝑦))} 𝑑𝑥 + 𝑓⁽𝑢⁽𝑦⁾, 𝑦⁾ ∙ 𝑢^′(𝑦⁾ − 𝑓⁽𝑤⁽𝑦⁾, 𝑦⁾ ∙ 𝑤^′(𝑦⁾.

𝜕𝑦

𝑢⁽𝑦⁾

𝜕𝑦

Maple muhitida funksiyaning hosilasini topish uchun

• 
  diff(f(x),x);
  

buyrug’idan foydalaniladi. Bu buyruqdan so’ng ekranga f(x) funksiyaning x bo’yicha hosilasi chop etiladi.
Misollar bilan tanishamiz:

• restart;

• 
  Int(exp(y*sqrt(xy)),x=sin(y)..cos(y));
  

cos( y )

• value(%);




sin( y )
_e( y xy ) _dx

• diff(x^2-3*x,x);

• 
  diff(sin(x^2),x);
  

e^{( y xy )} ( cos( y)  sin( y ))


2 x  3

2 cos( x² ) x

• F(y):=Int(exp(y*sqrt(1-x^2)),x=sin(y)..cos(y));

cos( y )

• diff(F(y),y);

F( y ) := ^



sin( y )

cos( y )
_
_
_e( y
^{1  x}2 ⁾ dx  sin( y ) e^{( y}
1  cos( y )² ) _{ cos( y ) e}( y
1  sin( y )² )

sin( y )

• F(y):=Int(ln(1+y*x)/x,x=0..y);

y
_{F( y ) :=} ^_ ln( 1  y x ) _dx

• diff(F(y),y);




0
y
_{ 1}

x

ln( 1  y² )

• value(%);

_ 1  y x ^{dx } y


0


ln( 1  y² )

• Int(x+y*x^2,x=y-1..y);

• 
  diff(%,y$2);
  

2
y


_
y  1
y
x  y x² dx

• Int(sin(x*y),x=0..1);

• 
  diff(%,y);
  

4 y²  2 ( y  1)²  2 y ( y  1)

1


^_ sin( y x ) dx
0
1


^_ cos( y x ) x dx
0

• Diff(Int(ln(x+y)*y^2,x=-1..y),y)=diff(Int(ln(x+y)*y^2,x=-1..y),y);

y



y

2
^ ln( x  y ) y² dx   ^y  2 ln( x  y ) y dx  ln( 2 y ) y²

y ^
-1
_ x  y


-1

• Diff(Int(x*y-y^2,x=0..1),y)=diff(Int(x*y-y^2,x=0..1),y);

_ 1 ₁
^_ y x  y² dx  ^_ x  2 y dx




y ^
₀ 0

Maple dasturida ham xosmas integrallarni hisoblash imkoniyati mavjud bo’lib, uni ham int funksiyasi orqali topiladi. Buni quyida keltirilgan misollar orqali mustaqil tahlil qiling.

>restart;

>Int(1/(x^2+6*x+12),x=-infinity..infinity);

_^ 1 _dx


^ x²  6 x  12



>value(%);

1 _
3

>Int(cos(a*x)/(1+x^2),x=-

infinity..+infinity)=int(cos(a*x)/(1+x^2),x=-infinity..infinity);

^{ cos( a x )} dx  signum( a )  sinh( a )


^ 1  x²



>convert(int(cos(a*x)/(1+x^2),x=-infinity..infinity),exp);

signum( a )  _^{ 1} e^a  ¹

1 _

_{ 2}

>int(sin(x^2),x=-infinity..infinity);

>int(x/(exp(x)-1),x=0..infinity);
1 _2
6

>int(exp(-x^2)*ln(x),x=0..infinity);

_{2 e}a _

  

>int(sin(a*x)/x,x=0..infinity);

ln( 2 )

¹ signum( a ) 
2

>Int(sin(a*x)/x^s,x=0..infinity)=int(sin(a*x)/x^s,x=0..infinity);

_ ₂^{( s )} _a^{( s  1 )} _{ 1 } ¹ _{s }

_ sin( a x ) _{dx 
 x}s
_
^_ ¹  ¹ s ^_
2 _

₀

>int(ln(1/x)^4,x=0..1);

_{ 2 2} _
24

>Int(sin(a*x)/x*ln(x),x=0..infinity)=int(sin(a*x)/x*ln(x),x=0..infi nity);


sin( a x ) ln( x ) _dx 1 1


_{ x}   ₂  ln( a )  ₂  
0

>Int(ln(x)/(x+1),x=0..1)=int(ln(x)/(x+1),x=0..1);

1
^_ ln( x ) _{dx  } 1 _2


_ x  1 12
0

>int(1/sqrt(1-x^a),x=0..1);

_ ¹ 
^ ^{a }
a ^_ ¹  ^{1 }_



>int(1/sqrt(2-x),x=0..2);

>int(1/sqrt(1-x^2),x=-1..1);

>int(1/x,x=0..1);

_{ 2}


2




_a _

Bundan tashqari xosmas integrallarda ham o’zgaruvchilarni almashtirish va bo’laklab integrallash imkoniyatlari Maple dasturida mavjud bo’lib, bu amallar ham changevar va intparts funksiyalari orqali amalga oshiriladi.

>restart;

>with(student):

>changevar(x-1=u,Int(1/sqrt(1-x^2),x=-1..1),u);

0
^_ 1 _du

^-2

>Int(1/sqrt((x-1)*(2-x)),x=1..2);

2
^_ 1 _dx

₁

>changevar(x=cos(phi)^2+2*sin(phi)^2,%,phi) ;

0
_
 2 d
^ 1/2 

>simplify(%,trig);

0




^ 2 d

 1/2 

>changevar(x^2=t,Int(sin(x^2),x=0..infinity),t);


_^ 1 sin( t ) _dt


_ 2
0

>intparts(Int(exp(-2*x)*cos(3*x),x=0..infinity),cos(3*x));



_{1 } _
³ sin( 3 x ) e^{( 2 x )} dx


2 _ 2
0

>Int(arctan(x)/(1+x^2)^1.5,x=1..infinity);



>intparts(%,arctan(x));

^ arctan( x )


dx
 ( 1  x² )1.5
₁


^ ^x

1.015435960 _ dx
₁
Maple muhitida 𝑦 = 𝑓⁽𝑥⁾ funksiyaning uzilish nuqtalarini aniqlash uchun
discont(f(x),x) buyrug’idan foydalaniladi.

• f(x):=1/((x-1)*(x-2)*(x-3));

• 
  discont(f(x),x);
  

• discont(tan(x),x);

• 
  discont(floor(x),x);
  

f( x ) :=
1

( x  1 ) ( x  2 ) ( x  3 )

{1, 2, 3}

{  _Z1~  ¹ }
2

{_Z2~}

Quyidagi misollarni mustaqil tahlil qiling:

>restart;

>series(int(sin(x)/x,x=t..infinity),t=0,7);

¹   t 
2
¹ t³ 
18
1

600
t⁵  O( t⁷ )

>Limit(Int(1/(1+x^n),x=0..infinity),n=infinity)=limit(int(1/(1+x^n)

,x=0..infinity),n=infinity);

lim
n  


^ ¹ dx  1


^ 1  xⁿ


0

>discont(int(sin((1-a^2)*x)/x,x=0..infinity),a);

{-1, 1 }

5 – 8 – mavzularni takrorlashga oid savol va topshiriqlar

1. 
   Funksiyaning aniq integralini Maple dasturida hisoblashdan maqsad nima?
   
2. 
   Maple dasturida funksiyaning aniq integrali qaysi buyruq yordamida hisoblanadi? Misollar bilan.
   
3. 
   rightsum, leftsum, middlesum funsiyalarining vazifasini ayting. Misollar bilan.
   
4. 
   Funksiya grafigini biror oraliqda to’g’ri to’rtburchaklar bilan qoplash uchun Maple dasturida qaysi buyruqlardan foydalanamiz? Misollar bilan.
   
5. 
   Maple dasturida funksiyaning aniq integralini bo’laklab integrallash orqali hisoblash uchun qaysi buyruqdan foydalanamiz? Misollar bilan.
   
6. 
   Maple dasturida funksiyaning aniq integralini o’zgaruvchini almashtirish orqali hisoblash uchun qaysi buyruqdan foydalanamiz?
   
7.