اﻟﺘﻜﺎﻣـــــﻞ
اﻟﺜﺎﻧﻴﺔ ﺳﻠﻚ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
-Iﺗﻜﺎﻣﻞ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل -1ﺗﻌﺮﻳﻒ و ﺗﺮﻣﻴﺰ ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iو aو bﻋﻨﺼﺮﻳﻦ ﻣﻦ . I إذا آﺎﻧﺖ Fو Gداﻟﺘﻴﻦ أﺻﻠﻴﺘﻴﻦ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ Iﻓﺎن ).F(b)-F(a)=G(b)-G(a أي أن اﻟﻌﺪد اﻟﺤﻘﻴﻘﻲ ) F(b)-F(aﻏﻴﺮ ﻣﺮﺗﺒﻂ ﺑﺎﺧﺘﻴﺎر اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ .F ﺗﻌﺮﻳﻒ ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iو aو bﻋﻨﺼﺮﻳﻦ ﻣﻦ .I اﻟﻌﺪد اﻟﺤﻘﻴﻘﻲ) F(b)-F(aﺣﻴﺚ Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ , Iﻳﺴﻤﻰ ﺗﻜﺎﻣﻞ اﻟﺪاﻟﺔ وﻳﻜﺘﺐ
b
∫a f ( x )dx
وﻳﻘﺮأ ﻣﺠﻤﻮع f ( x ) dxﻣﻦ aإﻟﻰ bأو ﺗﻜﺎﻣﻞ ﻣﻦ aإﻟﻰ bﻟـ . f ( x ) dx b
∫a f ( x )dx
aو bﻳﺴﻤﻴﺎ ﻣﺤﺪا اﻟﺘﻜﺎﻣﻞ b
ﻓﻲ اﻟﻜﺘﺎﺑﺔ ∫a f ( x )dx b b b f x dx = f t dt = ( ) ( ) ∫a ∫a ∫a f (u )du = ........
ﻳﻤﻜﻦ ﺗﻌﻮﻳﺾ xﺑﺄي ﺣﺮف ﺁﺧﺮ ،ﺑﻤﻌﻨﻰ أن
b
b ∫a f ( x )dx = F ( x )a
ﻣﻦ أﺟﻞ ﺗﺒﺴﻴﻂ اﻟﻜﺘﺎﺑﺔ ) F(b)-F(aﻧﻜﺘﺒﻬﺎ ﻋﻠﻰ اﻟﺸﻜﻞ أﻣﺜﻠﺔ *
ﻧﺤﺴﺐ
1 اﻟﺪاﻟﺔ x اذن *
1
2
∫1 x dx → xﻣﺘﺼﻠﺔ ﻋﻠﻰ ] [1; 2و داﻟﺔ أﺻﻠﻴﺔ ﻟﻬﺎ هﻲ x → ln x
1
2
∫1 x dx = [ln x ]1 = ln 2 2
0
∫ 2 cos xdx
أﺣﺴﺐ
π
1 ∫−1 x 2 + 1dx 1
;
1 dx cos 2 x
;
π
∫
4 0
-2ﺧﺎﺻﻴﺎت أ -ﺧﺎﺻﻴﺎت ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iو aو bو cﻋﻨﺎﺻﺮ ﻣﻦI a
a
* f ( x ) dx = − ∫ f ( x ) dx
* ∫a f ( x )dx = 0 b c b f x dx = f x dx + ( ) ( ) ∫a ∫a * ∫c f ( x )dx b
b
∫a
)ﻋﻼﻗﺔ ﺷﺎل(
أﻣﺜﻠﺔ 1
أﺣﺴﺐ I = ∫ x dx −1
1
fﻣﻦ aإﻟﻰ b
0
−1 2 1 2 x dx = x dx = − xdx + xdx = x + ∫−1 ∫−1 ∫−1 ∫0 2 2 x = 1 −1 0 0
1
ب( -ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iو aﻋﻨﺼﺮا ﻣﻦ I
1
1
1
→ ϕ :I x
x → ∫ f (t ) dt a
ﻟﺪﻳﻨﺎ ) ϕ ( x ) = F ( x ) − F ( a
∀x ∈ Iﺣﻴﺚ Fداﻟﺔ أﺻﻠﻴﺔ ﻟـ fﻋﻠﻰ .I ϕ ( a ) = 0أي أن ϕداﻟﺔ اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ Iاﻟﺘﻲ ﺗﻨﻌﺪم
اذن ϕﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ Iو ϕ ' = f ﻓﻲ a ﺧﺎﺻﻴﺔ
ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iو aﻋﻨﺼﺮا ﻣﻦ .I x
اﻟﺪاﻟﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ Iﺑﻤﺎ ﻳﻠﻲ x → ∫ f ( t ) dtهﻲ اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟـ fﻋﻠﻰ Iاﻟﺘﻲ ﺗﻨﻌﺪم ﻓﻲ a a
1 ﻧﻌﻠﻢ أن اﻟﺪاﻟﺔ x → ln xهﻲ اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟـ x
ﻣﺜﺎل
1 dt t
x
→ xﻋﻠﻰ [∞ ]0; +اﻟﺘﻲ ﺗﻨﻌﺪم ﻓﻲ. 1
[∞∀x ∈ ]0; +
∫ = ln x
1
ﺗﻤﺮﻳﻦ ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟـ fﻋﻠﻰ [∞ ]0; +اﻟﺘﻲ ﺗﻨﻌﺪم ﻓﻲ 2ﺣﻴﺚ
1 ln x x
= )f ( x
[∞∀x ∈ ]0; +
ج ( -ﺧﺎﺻﻴﺔ ﻟﺘﻜﻦ fو gداﻟﺘﻴﻦ ﻣﺘﺼﻠﺘﻴﻦ ﻋﻠﻰ ] [ a; bو λﻋﺪد ﺣﻘﻴﻘﻲ ﺛﺎﺑﺖ
∫a ( f ( x ) + g ( x ) )dx = ∫a f ( x )dx + ∫a g ( x )dx π 1 4 2 ﺣﺪد ) ∫ cos xdx ; ∫ ( x − 3x + 1) dxﻳﻤﻜﻦ اﺧﻄﺎط x 0 0 b
ﺗﻤﺮﻳﻦ ﺗﻤﺮﻳﻦ
b
cos x dx sin x + cos x
ﻧﻌﺘﺒﺮ
أﺣﺴﺐ I + J
b
π
b
b
( cos 4
π
I =∫4 0
I−J
∫a ( λ f ( x ) )dx = λ ∫a f ( x )dx
و اﺳﺘﻨﺘﺞ ; I
sin x J =∫4 dx 0 sin x + cos x J
b
د اﻟﺘﺄوﺑﻞ اﻟﻬﻨﺪﺳﻲ ﻟﻠﻌﺪد ∫a f ( x ) dx
b
A ( f ) = ∫ f ( x)dx a
ﺧﺎﺻﻴﺔ إذا آﺎﻧﺖ fداﻟﺔ ﻣﺘﺼﻠﺔ و ﻣﻮﺟﺒﺔ ﻋﻠﻰ ] ( a ≺ b ) [ a; bﻓﺎن ﻣﺴﺎﺣﺔ اﻟﺤﻴﺰ اﻟﻤﺤﺼﻮر ﺑﻴﻦ ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ f و ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ و اﻟﻤﺴﺘﻘﻴﻤﻴﻦ اﻟﻤﻌﺮﻓﺘﻴﻦ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ x = aو x = bهﻲ b
) = ∫a f ( x )dx ﻣﻼﺣﻈﺔ ﺗﻤﺮﻳﻦ
A ( fﺑﻮﺣﺪة ﻗﻴﺎس اﻟﻤﺴﺎﺣﺎت
إذا آﺎن اﻟﻤﺴﺘﻮى ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪﻳﻦ ﻓﺎن وﺣﺪة ﻗﻴﺎس اﻟﻤﺴﺎﺣﺔ هﻲ ﻣﺴﺎﺣﺔ اﻟﻤﺮﺑﻊ OIJK
1 ﻧﻌﺘﺒﺮ x2
= )f ( x
2
أﻧﺸﺊ C f
)
= 1cm
j = 2cm
(i
أﺣﺴﺐ ﺑـ cm 2ﻣﺴﺎﺣﺔ اﻟﺤﻴﺰ اﻟﻤﺤﺼﻮر ﺑﻴﻦ C fو ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ و اﻟﻤﺴﺘﻘﻴﻤﻴﻦ اﻟﻤﻌﺮﻓﻴﻦ ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ .x = 3 ; x =1 -IIﺗﻘﻨﻴﺎت ﺣﺴﺎب اﻟﺘﻜﺎﻣﻼت -1اﻻﺳﺘﻌﻤﺎل اﻟﻤﺒﺎﺷﺮ ﻟﺪوال اﻷﺻﻠﻴﺔ أﻣﺜﻠﺔ
( ln x )2 dx
e
( ln x )2
∫1 x x 2 e e ) e ( ln x 1 3 1 3 1 3 1 2 و ﻧﻌﻠﻢ أن اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟـ u 'uهﻲ 3 uإذن ∫1 x dx = 3 u ( x )1 = 3 ln x 1 = 3 1 2 2 e −x 2 ﻳﻜﺘﺐ ﻋﻠﻰ ﺷﻜﻞ أن ﻧﻼﺣﻆ اﻟﺘﺤﻮﻳﻞ ﺬا ﺑﻬ = ﻟﺪﻳﻨﺎ 2 * أﺣﺴﺐ ∫0 e x + 1dx 1+ex ex +1 1 + e −x 1 1 1 2 'u ﺣﻴﺚ u ( x ) = 1 + e − xإذن − ln 1 + e − x dx u x −2 2ln = − = ( ) ∫0 e x + 1 0 0 u أﺣﺴﺐ
*
ﻧﻼﺣﻆ أن
)
ﺗﻤﺮﻳﻦ
ﺣﺪد xdx
-1
3
ﻋﻠﻰ ﺷﻜﻞ
2
u 'uﺣﻴﺚ u ( x ) = ln x
(
π
∫04 sin
2x 4 + x 2 + x − 1 b c -2أ -أوﺟﺪ aو bو cﺣﻴﺚ ∀x ≠ 0 = + + ax x x 2 +1 x3 +x 4 2 3 2x + x + x − 1 ب -اﺳﺘﻨﺘﺞ ﻗﻴﻤﺔ dx ∫1 x3 +x '1 u 1 ﺣﻴﺚ uداﻟﺔ ﻳﺠﻴﺐ ﺗﺤﺪﻳﺪهﺎ . ﺷﻜﻞ ﻋﻠﻰ ﻳﻜﺘﺐ -3ﺑﻴﻦ أن اﻟﺘﻌﺒﻴﺮ 2u 2 +1 x 2 − 2x + 5 1+ 2 3 1 اﺳﺘﻨﺘﺞ ﻗﻴﻤﺔ ∫1 x 2 − 2x + 5 dx 1 1 e2 1 1 1 x ; = dx ∫e x ln x -4أﺣﺴﺐ ∫0 ( x + 1)( x + 2 ) dx ln ln x x x -2اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء a ; b ﻋﻠﻰ ﻣﺘﺼﻠﺘﻴﻦ g ' و f ' ﺑﺤﻴﺚ a ; b ﻋﻠﻰ ﻟﻼﺷﻨﻘﺎق ﻗﺎﺑﻠﺘﻴﻦ داﻟﺘﻴﻦ ﻟﺘﻜﻦ fو g ] [ ] [ ﻧﻌﻠﻢ أن
) ( fg ) ' ( x ) = f ' ( x ) g ( x ) + f ( x ) g ' ( x ) f ' ( x ) g ( x ) = ( fg ) ' ( x ) − f ( x ) g ' ( x
] ∀x ∈ [a;b ] ∀x ∈ [a;b
ﺧﺎﺻﻴﺔ b
b
b ∫a f '( x ) g ( x )dx = ( fg )( x )a − ∫a f ( x ) g '( x )dx
π
ﻣﺜﺎل
أﺣﺴﺐ
∫02 x cos xdx
ﻧﻀﻊ ; u ' ( x ) = cos x
وﻣﻨﻪ v ' ( x ) = 1 ; u ( x ) = sin x
3
v (x ) = x
π
π
π
π
π
π
∫02 x cos xdx = [ x sin x ]02 − ∫02 sin xdx = [ x sin x ]02 − [ − cos x ]02 = 2 − 1 π
π
K = ∫ 2 e x sin xdx
; J = ∫ x 2 sin xdx
0
0
إذن ﺗﻤﺮﻳﻦ e
; I = ∫ ln xdx أﺣﺴﺐ 1
اﻟﺤﻞ π
π
π
π
K = e x sin x 2 − ∫ 2 e x cos xdx = e x sin x 2 − e x cos x 2 − K 0 0 0 0 π π 1 x x 2 K = e sin x − e cos x 2 = ......... 0 0 2
x +2 ∫0 x + 1 dx x f (x ) = ﺣﻴﺚ cos 2 x 1
ln
1
∫0
3
∫0 ( x
x x + 3dx
2
− 1)e 2 x dx
∫1 x
2
ln xdx
أﺣﺴﺐ-1 ﺗﻤﺮﻳﻦ
π π − 2 ; 2 ﻋﻠﻰf ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء أوﺟﺪ اﻟﺪوال اﻷﺻﻠﻴﺔ ﻟـ-2 (J =
x
∫0
x
e t sin 2 tdt ) ﻳﻤﻜﻦ اﻋﺘﺒﺎرI = ∫ e t cos 2 tdt 0
أﺣﺴﺐ-3 اﻟﺘﻜﺎﻣﻞ و اﻟﺘﺮﺗﻴﺐ-III
ﻣﻘﺎرﻧﺔ ﺗﻜﺎﻣﻠﻴﻦ-1 [a;b] ﻋﻠﻰf داﻟﺔ أﺻﻠﻴﺔ ﻟـF [ وa;b] داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰf ( ﻟﺘﻜﻦa b
∀x ∈ [a;b ] F ' ( x ) = f ( x )
[ a; b ] ﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ
∫a f ( x )dx = F (b ) − F (a )
F [ ﻓﺎنa; b ] ﻣﻮﺟﺒﺔ ﻋﻠﻰf إذا آﺎﻧﺖ
b
∫a f ( x )dx ≥ 0 ادن
F ( a ) ≤ F ( b ) ﻓﺎنa ≤ b وﺣﻴﺚ أن
( a ≤ b ) [ a; b ]
ﺧﺎﺻﻴﺔ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰf ﻟﺘﻜﻦ
b
∫a f ( x )dx ≥ 0 [ ﻓﺎنa; b] ﻣﻮﺟﺒﺔ ﻋﻠﻰ ( a ≤ b ) [ a; b ] b
∫a
f إذا آﺎﻧﺖ
( ﺧﺎﺻﻴﺔb داﻟﺘﻴﻦ ﻣﺘﺼﻠﺘﻴﻦ ﻋﻠﻰg وf ﻟﺘﻜﻦ
b
f ( x ) dx ≤ ∫ g ( x ) dx [ ﻓﺎنa; b ] ﻋﻠﻰf ≤ g إذا آﺎﻧﺖ a ﻣﺜﺎل 2
x dx ﻧﺆ ﻃﺮ 01+ x
I =∫ 1x
∫0
2
2
1
1
x2 x2 1≤1+ x ≤ 2 ⇔ ≤ ≤x2 2 1+ x
dx ≤ I ≤ ∫ x dx ∀ وﻣﻨﻪx ∈ [ 0;1] 2
0
1 1 ≤I≤ 6 3
4
ﻟﺪﻳﻨﺎ إذن ( ﺧﺎﺻﻴﺎتc
أ-
ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ] ( a ≤ b ) [ a; b
إذا آﺎﻧﺖ f
b
ﺳﺎﻟﺒﺔ ﻋﻠﻰ ] [ a; bﻓﺎن ∫a f ( x )dx ≤ 0 b
b
∫a f ( x )dx ≤ ∫a f ( x ) dx
ب-
ج -ﻟﺘﻜﻦ Mاﻟﻘﻴﻤﺔ اﻟﻘﺼﻮﻳﺔ و mاﻟﻘﻴﻤﺔ اﻟﺪﻧﻮﻳﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ
] [ a; b
b
) m (b − a ) ≤ ∫ f ( x ) dx ≤ M (b − a a
ﻣﻼﺣﻈﺔ b
) = ∫a f ( x )dx
إذا آﺎﻧﺖ fﻣﻮﺟﺒﺔ ﻋﻠﻰ ] [ a; bﻓﺎن اﻟﻤﺴﺎﺣﺔ
A ( fﻓﻲ ﻣﻌﻠﻢ م.م ﻣﺤﺼﻮرة ﺑﻴﻦ
ﻣﺴﺎﺣﺘﻲ اﻟﻤﺴﺘﻄﻴﻞ اﻟﺬي ﺑﻌﺪﻳﻪ Mو ) ( b − aو اﻟﻤﺴﺘﻄﻴﻞ اﻟﺬي ﺑﻌﺪﻳﻪ mو ) . ( b − a
b
A ( f ) = ∫ f ( x)dx a
ﻣﺜﺎل
ﻧﻌﺘﺒﺮ dx
1 2
x 1+ x
3
∫= I
1
ﻧﺒﻴﻦ أن 0 ≤ I ≤ 2
2 1 → xﻣﻮﺟﺒﺔ و ﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ [∞ ]0; +وﻣﻨﻪ اﻟﺪاﻟﺔ 2 x 1 + x2
= )sup f ( x ) = f (1 ]x ∈[1;3
2 اذن 2 -2اﻟﻘﻴﻤﺔ اﻟﻤﺘﻮﺳﻄﺔ ﻟﺪاﻟﺔ ﻣﺘﺼﻠﺔ ﻓﻲ ﻗﻄﻌﺔ ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ] ( a ≺ b ) [ a; bو Mاﻟﻘﻴﻤﺔ اﻟﻘﺼﻮﻳﺔ و mاﻟﻘﻴﻤﺔ اﻟﺪﻧﻮﻳﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ ][ a; b )0 ≤ I ≤ ( 3 − 1
1 b إذن f ( x ) dx ≤ M b − a ∫a 1 b = )f (c ﺣﻴﺚ f ( x ) dx b − a ∫a ﺧﺎﺻﻴﺔ و ﺗﻌﺮﻳﻒ ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ] ( a ≠ b ) [ a; b ≤m
1 b اﻟﻌﺪد اﻟﺤﻘﻴﻘﻲ f ( x ) dx b − a ∫a
وﻣﻨﻪ ﺣﺴﺐ ﻣﺒﺮهﻨﺔ اﻟﻘﻴﻤﺔ اﻟﻮﺳﻄﻴﺔ ﻳﻮﺟﺪ ﻋﻠﻰ اﻷﻗﻞ cﻓﻲ
= µﻳﺴﻤﻰ اﻟﻘﻴﻤﺔ اﻟﻤﺘﻮﺳﻄﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ ] . [ a; b
1 b ﻳﻮﺟﺪ ﻋﻠﻰ اﻷﻗﻞ cﻓﻲ ] [ a; bﺣﻴﺚ f ( x ) dx b − a ∫a ﻣﻼﺣﻈﺔ
= )f (c
إذا آﺎﻧﺖ fﻣﻮﺟﺒﺔ ﻋﻠﻰ ] [ a; bﻓﺎن اﻟﻤﺴﺎﺣﺔ A ( f ) = ∫ f ( x ) dxﻓﻲ ﻣﻌﻠﻢ م.م هﻲ ﻣﺴﺎﺣﺔ b
a
اﻟﻤﺴﺘﻄﻴﻞ اﻟﺬي ﺑﻌﺪاﻩ ) ( b − aو ) . f ( c
5
] [ a; b
ﺗﻤﺮﻳﻦ -1أﺣﺴﺐ اﻟﻘﻴﻤﺔ اﻟﻤﺘﻮﺳﻄﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ Iﻓﻲ اﻟﺤﺎﻟﺘﻴﻦ اﻟﺘﺎﻟﻴﺘﻴﻦ
x3 + 5 x 2 + x + 3 ](b ; I = [ −1;0 (a f ( x ) = ( x − 1) e x x +1 -2أﻃﺮ اﻟﺪاﻟﺔ fﻋﻠﻰ ] [ 0;1ﺣﻴﺚ f ( x ) = arctan x 1 ﻟﺪﻳﻨﺎ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ] [ 0;1و 1 + x2
اﻟﺠﻮاب ﻋﻦ اﻟﺴﺆال2
1 ≤ f '( x) ≤ 1 2
x x 1 ] ∀x ∈ [ 0;1ادن dt ≤ ∫ f ' (t ) dt ≤ ∫ dt 0 0 2
= )f ( x
] ∀x ∈ [ 0;1و ﻣﻨﻪ
= )f '( x
x
]I = [ 0;1
x ]≤ f ( x ) ≤ x ∀x ∈ [ 0;1 2
∫0
-IVﺣﺴﺎب اﻟﻤﺴﺎﺣﺎت -1ﺣﺴﺎب اﻟﻤﺴﺎﺣﺎت اﻟﻬﻨﺪﺳﻴﺔ اﻟﻤﺴﺘﻮى ﻣﻨﺴﻮب إﻟﻰ م.م.م
) ( o; i ; j
ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ] [ a; bو C fﻣﻨﺤﻨﺎهﺎ و ) ∆ ( fاﻟﺤﻴﺰ اﻟﻤﺤﺼﻮر ﺑﻴﻦ C fو ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ
( ∆1 ) : x = a
و اﻟﻤﺴﺘﻘﻴﻤﻴﻦ
( ∆2 ) : x = b
b
b
a
a
A ( f ) = ∫ − f ( x)dx = ∫ f ( x) dx ) A( f
*إذ ا آﺎﻧﺖ f
b
ﻣﻮﺟﺒﺔ ﻋﻠﻰ ] [ a; bﻓﺎن ﻣﺴﺎﺣﺔ ) ∆ ( fهﻲ ∫a f ( x )dx
ﺑﻮﺣﺪة ﻗﻴﺎس اﻟﻤﺴﺎﺣﺎت
*إذا آﺎن آﺎﻧﺖ fﺳﺎﻟﺒﺔ ﻋﻠﻰ ] [ a; bﻣﺴﺎﺣﺔ هﻲ ﻣﺴﺎﺣﺔ ) ∆ ( − f
(x ) dx
f
b
∫a
=
( x )d x
−f
b
) = ∫a
A (f
* إذا آﺎﻧﺖ fﺗﻐﻴﺮ إﺷﺎرﺗﻬﺎ ﻋﻠﻰ ] [ a; bﻣﺜﻼ ﻳﻮﺟﺪ cﻣﻦ ] [ a; bﺣﻴﺚ fﻣﻮﺟﺒﺔ ﻋﻠﻰ ] [ a; cو ﺳﺎﻟﺒﺔ ﻋﻠﻰ
][c; b اﻟﺤﻴﺰ ) ∆ ( fﻋﻠﻰ ] [ a; bهﻮ اﺗﺤﺎد ) ∆ ( fﻋﻠﻰ ] [ a; cو ) ∆ ( fﻋﻠﻰ b
b
b
c
][c; b c
) = ∫a f ( x ) dx + ∫c −f ( x )dx = ∫a f ( x ) dx + ∫c f ( x ) dx = ∫a f ( x ) dx
A (f
ﺧﺎﺻﻴﺔ اﻟﻤﺴﺘﻮى ﻣﻨﺴﻮب اﻟﻰ م.م.م ) ( o; i ; j ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ] [ a; bو C fﻣﻨﺤﻨﺎهﺎ و ) ∆ ( fاﻟﺤﻴﺰ اﻟﻤﺤﺼﻮر ﺑﻴﻦ C fو ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ
6
( ∆1 ) : x = a
و اﻟﻤﺴﺘﻘﻴﻤﻴﻦ
ﻣﺴﺎﺣﺔ اﻟﺤﻴﺰ ) ∆ ( fهﻮ
( ∆2 ) : x = b b
∫a f ( x ) dx
ﺑﻮﺣﺪة ﻗﻴﺎس اﻟﻤﺴﺎﺣﺔ
اﺻﻄﻼﺣﺎت اﻟﻌﺪد اﻟﻤﻮﺟﺐ اﻟﻌﺪد اﻟﺤﻘﻴﻘﻲ
b
∫ f ( x ) dx ∫ f ( x ) dxﻳﺴﻤﻰ اﻟﻤﺴﺎﺣﺔ اﻟﺠﺒﺮﻳﺔ ﻟﻠﺤﻴﺰ ) . ∆ ( f a
ﻳﺴﻤﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻬﻨﺪﺳﻴﺔ ﻟﻠﺤﻴﺰ ) . ∆ ( f
b
a
ﻣﺜﺎل ﻧﻌﺘﺒﺮ f ( x ) = x − 1 3
ﺣﺪد ﻣﺴﺎﺣﺔ اﻟﺤﻴﺰ اﻟﻤﺤﺼﻮر ﺑﻴﻦ اﻟﻤﻨﺤﻨﻰ C fو ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ و اﻟﻤﺴﺘﻘﻴﻤﻴﻦ ذا اﻟﻤﻌﺎدﻟﺘﻴﻦ x=2 ; x=0 2
A = ∫ f ( x ) dx 0
) ) j
− 1 dx
(x
1
×
(u = i
3
2
)
(
1
∫ A = ∫ 1 − x3 dx + 0
7 A= u 2
ﻣﺴﺎﺣﺔ ﺣﻴﺰ ﻣﺤﺼﻮر ﺑﻴﻦ ﻣﻨﺤﻨﻴﻴﻦ -2 ﻟﺘﻜﻦ fو gداﻟﺘﻴﻦ ﻣﺘﺼﻠﺘﻴﻦ ﻋﻠﻰ ] [ a; b
( ∆1 ) : x = a
و ∆ هﻮ اﻟﺤﻴﺰ اﻟﻤﺤﺼﻮر ﺑﻴﻦ C fو Cgو اﻟﻤﺴﺘﻘﻴﻤﻴﻦ
b
∫a f ( x ) − g ( x ) dx
7
( ∆2 ) : x = b
ﻓﻲ م.م.م
) ( o; i ; j
إذا آﺎن f ≥ g ≥ 0ﻓﺎن ) A ( ∆ ) = A ( f ) − A ( g
A ( ∆ ) = ∫ f ( x ) dx − ∫ g ( x ) dx = ∫ ( f ( x ) − g ( x ) ) dx = ∫ f ( x ) − g ( x ) dx b
b
b
b
a
a
a
a
إذا آﺎﻧﺖ f ≤ gو آﻴﻔﻤﺎ آﺎﻧﺖ إﺷﺎرﺗﻲ fو gو ﺑﺈﺗﺒﺎع ﻧﻔﺲ اﻟﻄﺮﻳﻘﺔ ﻧﺤﺼﻞ ﻋﻠﻰ أن b
A ( ∆ ) = ∫ f ( x ) − g ( x ) dx a
ﺧﺎﺻﻴﺔ
] [ a; b
ﻟﺘﻜﻦ fو gداﻟﺘﻴﻦ ﻣﺘﺼﻠﺘﻴﻦ ﻋﻠﻰ
ﻣﺴﺎﺣﺔ اﻟﺤﻴﺰ ∆ اﻟﻤﺤﺼﻮر ﺑﻴﻦ C fو Cgو اﻟﻤﺴﺘﻘﻴﻤﻴﻦ هﻲ
b
∫a f ( x ) − g ( x ) dx
=a
( ∆1 ) : x
=b
( ∆2 ) : x
= ) ∆ ( Aوﺣﺪة ﻗﻴﺎس اﻟﻤﺴﺎﺣﺎت
ﻣﻼﺣﻈﺔ c
∫a f ( x ) − g ( x ) dx
b
∫c g ( x ) − f ( x ) dx
( g ( x ) − f ( x ) )dx -Vﺣﺴﺎب اﻟﺤﺠﻮم ﻓﻲ اﻟﻔﻀﺎء
)
∫ A ( ∆ ) = ∫ ( f ( x ) − g ( x ) ) dx +
b
c
c
a
(
اﻟﻔﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ م.م o; i ; j ; kﻧﻔﺘﺮض أن وﺣﺪة ﻗﻴﺎس اﻟﺤﺠﻢ هﻲ ﺣﺠﻢ اﻟﻤﻜﻌﺐ اﻟﺬي ﻃﻮل ﺣﺮﻓﻪ
i -1ﺣﺠﻢ ﻣﺠﺴﻢ ﻓﻲ اﻟﻔﻀﺎء ﻟﻴﻜﻦ Sﻣﺠﺴﻤﺎ ﻣﺤﺼﻮرا ﺑﻴﻦ اﻟﻤﺴﺘﻮﻳﻴﻦ اﻟﻤﻌﺮﻓﻴﻦ ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ z = aو z = b ﻧﺮﻣﺰ ﺑـ ) S ( tإﻟﻰ ﻣﺴﺎﺣﺔ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) M ( x; y; zﻣﻦ Sﺣﻴﺚ z = tو ﺑﺎﻟﺮﻣﺰ ) V ( tإﻟﻰ ﺣﺠﻢ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ﻣﻦ Sاﻟﻤﺤﺼﻮر ﺑﻴﻦ اﻟﻤﺴﺘﻮﻳﻴﻦ z = t ; z = a ﻟﻴﻜﻦ t0ﻣﻦ ] [ a; bو hﻋﺪدا ﻣﻮﺟﺒﺎ ﺣﻴﺚ ] t0 + h ∈ [ a; b
8
ﺣﺠﻢ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) M ( x; y; zﻣﻦ Sاﻟﻤﺤﺼﻮرة ﺑﻴﻦ z = t0و z = t0 + hهﻮ ) V ( t0 + h ) − V ( t0
وﻣﻦ ﺟﻬﺔ ﺛﺎﻧﻴﺔ هﺬا اﻟﺤﺠﻢ ﻣﺤﺼﻮر ﺑﻴﻦ ﺣﺠﻤﻲ اﻷﺳﻄﻮاﻧﺘﻴﻦ اﻟﺘﻲ ارﺗﻔﺎﻋﻬﻤﺎ hو ﻣﺴﺎﺣﺘﺎ ﻗﺎﻋﺪﺗﻴﻬﻤﺎ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ ) S ( t0و ) S ( t0 + h إذا اﻓﺘﺮﺿﻨﺎ أن ) S ( t0 ) ≤ S ( t0 + hﻓﺎن ) h ⋅ S ( t0 ) ≤ V ( t0 + h ) − V ( t0 ) ≤ h ⋅ S ( t0 + h
و ﻣﻨﻪ ) ≤ S ( t0 + h
) V ( t0 + h ) − V ( t0 h
≤ ) S ( t0
و إذا اﻓﺘﺮﺿﻨﺎ أن اﻟﺘﻄﺒﻴﻖ ) t → S ( tﻣﺘﺼﻞ ﻋﻠﻰ ] [ a; bﻓﺎن ) = S ( t0 إذن اﻟﺪاﻟﺔ ) t → V ( tﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ] [ a; bو ) V ' ( t ) = S ( t أي أن اﻟﺪاﻟﺔ ) t → V ( tداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ) t → S ( t t
و ﺑﻤﺎ أن V ( a ) = 0ﻓﺎن V ( t ) = ∫ S ( x ) dx a
) V ( t0 + h ) − V ( t0 h ] ∀t ∈ [ a; b
lim h →0
ﻋﻠﻰ ] [ a; b
] ∀t ∈ [ a; b
b
إذن ﺣﺠﻢ اﻟﻤﺠﺴﻢ Sهﻮ V = V ( b ) = ∫ S ( x ) dxوﺣﺪة ﻗﻴﺎس اﻟﺤﺠﻢ . a
ﺧﺎﺻﻴﺔ اﻟﻔﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ م.م ﻟﻴﻜﻦ Sﻣﺠﺴﻤﺎ ﻣﺤﺼﻮرا ﺑﻴﻦ اﻟﻤﺴﺘﻮﻳﻴﻦ اﻟﻤﻌﺮﻓﻴﻦ ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ z = aو z = b ﻧﺮﻣﺰ ﺑـ ) S ( tاﻟﻰ ﻣﺴﺎﺣﺔ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) M ( x; y; zﻣﻦ Sﺣﻴﺚ z = t إذا آﺎن أن اﻟﺘﻄﺒﻴﻖ ) t → S ( tﻣﺘﺼﻼ ﻋﻠﻰ ] [ a; bﻓﺎن ﺣﺠﻢ اﻟﻤﺠﺴﻢ Sهــﻮ V = ∫ S ( z ) dzوﺣﺪة ﻗﻴﺎس b
a
اﻟﺤﺠﻢ.
9
ﺗﻤﺮﻳﻦ أﺣﺴﺐ ﺣﺠﻢ اﻟﻔﻠﻜﺔ اﻟﺘﻲ ﻣﺮآﺰهﺎ Oو ﺷﻌﺎﻋﻬﺎ R اﻟﺤﻞ :ﻧﻔﺘﺮض أن اﻟﻔﻀﺎء ﻣﻨﺴﻮب م.م.م أﺻﻠﻪ . O اﻟﻔﻠﻜﺔ ﻣﺤﺼﻮرة ﺑﻴﻦ اﻟﻤﺴﺘﻮﻳﻴﻦ اﻟﻤﻌﺮﻓﻴﻦ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ z = − R ; z = R
ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) M ( x; y; zﻣﻦ اﻟﻔﻠﻜﺔ ﺣﻴﺚ z = t −R ≤ t ≤ R
و ﻣﺴﺎﺣﺘﻪ
)
هﻲ ﻗﺮص ﺷﻌﺎﻋﻪ 2
S (t ) = π ( R − t
R2 − t 2
2
ﺑﻤﺎ أن اﻟﺘﻄﺒﻴﻖ ) t → π ( R 2 − t 2ﻣﺘﺼﻠﺔ ﻋﻠﻰ
)
] [ − R; R
(
R 4 ﻓﺎن V = ∫ π R 2 − t 2 dt = π R 3 −R 3
-2ﺣﺠﻢ ﻣﺠﺴﻢ اﻟﺪوران ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ] [ a; bو C fﻣﻨﺤﻨﺎهﺎ ﻓﻲ م.م.م
) ( O; i ; j
إذا دار C fﺣﻮل اﻟﻤﺤﻮر ) ( O; iدورة آﺎﻣﻠﺔ ﻓﺎﻧﻪ ﻳﻮﻟﺪ ﻣﺠﺴﻤﺎ ﻳﺴﻤﻰ ﻣﺠﺴﻢ اﻟﺪوران
ﻓﻲ هﺬﻩ اﻟﺤﺎﻟﺔ ﻟﺪﻳﻨﺎ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) M ( x; y; zﻣﻦ اﻟﺠﺴﻢ ﺑﺤﻴﺚ x = tهﻲ ﻗﺮص ﻣﺴﺎﺣﺘﻪ
) (t
2
S (t ) = π f
اﻟﺘﻄﺒﻴﻖ
) (t
2
t → π fﻣﺘﺼﻠﺔ ﻋﻠﻰ
إذن ﺣﺠﻢ اﻟﻤﺠﺴﻢ اﻟﺪوراﻧﻲ هﻮ
] [ a; b
(t ) dt
2
b
V = ∫ πf a
ﺧﺎﺻﻴﺔ اﻟﻔﻀﺎء ﻣﻨﺴﻮب إﻟﻰ م.م.م أﺻﻠﻪ , oو fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ] [ a; b ﺣﺠﻢ ﻣﺠﺴﻢ اﻟﺪوران اﻟﻤﻮﻟﺪ ﻋﻦ دوران اﻟﻤﻨﺤﻨﻰ C fﺣﻮل اﻟﻤﺤﻮر ) ( OXهﻮ ﺑﻮﺣﺪة ﻗﻴﺎس اﻟﺤﺠﻢ .
10
(t ) dt
2
b
V = ∫ πf a
ﺗﻤﺮﻳﻦ 1 ﻧﻌﺘﺒﺮ f ( x ) = x ln x 2 أﻧﺸﺊ C fو ﺣﺪد ﺣﺠﻢ ﻣﺠﺴﻢ اﻟﺪوران اﻟﺬي ﻳﻮﻟﺪﻩ دوران اﻟﻤﻨﺤﻨﻰ C fﺣﻮل اﻟﻤﺤﻮر
11
) ( OXﻓﻲ اﻟﻤﺠﺎل
][1;e