اﻟﺘﻜﺎﻣـــــﻞ
اﻟﺜﺎﻧﻴﺔ ﺳﻠﻚ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
ﺗﻤﺎرﻳﻦ و ﺣﻠﻮل 4ﺗﻤﺮﻳﻦ
1 1 1 ﺗﺄآﺪ أن/ أ-1 − = t t + 2 t (t + 2)
dt
2
∫1 t ( t + 2 ) dt أﺣﺴﺐ/ب π
∫02 e π
x
sin xdx
أﺣﺴﺐ-2
π
J = ∫ x sin xdx ; I = ∫ 2 x 2 cos 2 xdx 2 0
2
2
ﻧﻀﻊ-3
0
J وI ﺛﻢ اﺳﺘﻨﺘﺞI − J وI + J أﺣﺴﺐ
1 1 1 ﺗﺄآﺪ أن/ أ-1 − = t t + 2 t (t + 2) 1 1 t +2−t 1 − = = t t + 2 t (t + 2) t (t + 2) 2
∫1 2
∫1
dt dt t (t + 2)
ﻧﺤﺴﺐ/ب
21 2 1 1 3 dt = ∫ − dt = ln t − ln ( t + 2 ) 1 = ln 2 − ln 4 + ln 3 = ln 1 t t (t + 2) t+2 2
π
A = ∫ 2 e x sin xdx 0
π
ﻧﺤﺴﺐ-2
π
π
π
A = 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 A = e x sin x 2 − e x cos x 2 0 0 2 π
= ......... π
J = ∫ 2 x sin xdx ; I = ∫ 2 x 2 cos 2 xdx 2
2
0
J +I =∫
π
2 x2 sin 2 0
xdx + ∫
π
2 x2 cos2 0
xdx = ∫
0
π
2 x2 0
( cos
2
)
π
π
1 3 2 2 x2 dx = 3 x 0 0
I − J ﻧﺤﺴﺐ
1
-3
I + J ﻧﺤﺴﺐ
x + sin x dx = ∫ 2
ﻧﻀﻊ
=
π3 24
I −J =∫
π
2 x2 cos2 0
xdx − ∫
π
2 x2 sin 2 0
π
π
π
(
)
xdx = ∫ 2 x2 cos2 x − sin2 x dx 0
π
sin2x 2 − 2 x sin2xdx I − J = ∫ 2 x2 cos2xdx = x2 0 2 0 ∫0 I − J = x
π
π
2 sin 2 x 2
π
cos 2 x cos 2 x 2 2 − − x 2 + ∫0 − 2 dx 0 0
2
π
−π I−J= 4
اذن
π
π
sin 2 x 2 cos 2 x 2 sin 2 x 2 I − J = x2 − −x + − 2 0 2 0 4 0 J وI ﻧﺴﺘﻨﺘﺞ
J=
π3 48
+
π 8
وI =
π3 48
−
(
π
f ( x) = e 1 − e x
lim
f ( x) x
x→+∞
وﻣﻨﻪ
8 x
π3
I−J=
و
24
−π 4
ﻟﺪﻳﻨﺎ 5ﺗﻤﺮﻳﻦ
)
;
I+J=
اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑـf ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ
lim f ( x ) ;
x→+∞
lim f ( x )
x→−∞
ﺣﺪد-1
C f و أﻧﺸﺊf و أﻋﻂ ﺟﺪول ﺗﻐﻴﺮاتf ' ( x ) أﺣﺴﺐ-2
و ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ و اﻟﻤﺴﺘﻘﻴﻤﻴﻦ اﻟﻤﻌﺮﻓﻴﻦC f اﻟﻤﺤﺼﻮر ﺑﻴﻦ (t
= e x ﻋﺪد ﺣﻘﻴﻘﻲ ﺳﺎﻟﺐ ) ﻳﻤﻜﻦ اﻋﺘﺒﺎرk
ﺣﻴﺚx = k
Ak
ﺣﺪد اﻟﻤﺴﺎﺣﺔ-3
; x = 0 ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ
lim Ak
ﺣﺪد-4
k →−∞
(
f ( x ) = ex 1 − ex f ( x) ; lim f ( x ) ; lim f ( x ) ﻧﺤﺪد x →+∞ x→+∞ x→−∞ x x x lim f ( x ) = lim e 1 − e = 0 ; lim f ( x ) = lim e x 1 − e x = −∞
)
lim
x →−∞
lim
x →+∞
f ( x) x
(
x→−∞
)
x→+∞
x→+∞
(
)
ex 1 − e x = −∞ ; = lim x→−∞ x
(
-4
)
C f و أﻧﺸﺊf و ﻧﻌﻄﻲ ﺟﺪول ﺗﻐﻴﺮاتf ' ( x ) أﻧﺴﺐ-5
(
f ' ( x ) = e x − e 2 x ' = e x − 2e 2 x = e x 1 − 2e x
)
ﺟﺪول اﻟﺘﻐﻴﺮات
2
x f '( x )
−∞
− ln 2 +
+∞
0
-
1 4
f 0
Ak
−∞
ﻧﺤﺪد اﻟﻤﺴﺎﺣﺔ-6
0
0
k
k
Ak = ∫ f ( x ) dx = ∫ e x − e 2 x dx 0
1 1 1 Ak = e x − e 2 x = − e k + e 2 k 2 2 k 2 lim Ak ﺣﺪد-4 k →−∞
1 1 1 − ek + e2k = k →−∞ 2 2 2
lim Ak = lim
k →−∞
ﺗﻤﺎرﻳﻦ
1ﺗﻤﺮﻳﻦ ∀x ∈\ −{−1;3}
b c −3x + 7x + 2 = a+ + 2 x +1 x − 3 x − 2x − 3 2
ﺣﻴﺚa ; b ; c ﺣﺪد-1
−3x 2 + 7 x + 2 أﺣﺴﺐ ∫0 x 2 − 2x − 3 dx π 4 2 1x + x +3 2 cos 4 xdx ∫ و أﺣﺴﺐ-2 dx ∫0 0 x2 + 1 e 2x − 1 e x − e −x ∀x ∈ \ = ﺑﻴﻦ أن-3 e 2x + 1 e x + e − x 2t x e −1 ∫0 e 2t + 1 dt أﺣﺴﺐ 2
ln2
2x ∫0 ( x +2)e dx
2 ∫0 x ln( x +1)dx 1 2
2ﺗﻤﺮﻳﻦ
π
;
∫02 x sinxdx
ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء أﺣﺴﺐ-1 π
∫0 e 3
x
sin xdx
و
-2ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟـ x → sin 3 xاﻟﺘﻲ ﺗﻨﻌﺪم ﻓﻲ 0ﻋﻠﻰ \ ﺛﻢ أﺣﺴﺐ dx ﺗﻤﺮﻳﻦ3 1
ﻧﻌﺘﺒﺮ I n = ∫ x n e x dx 0
-1أﺣﺴﺐ I1
-2ﺑﻴﻦ I n +1 = e − ( n + 1) I n
-3أﺣﺴﺐ I 2
*
` ∈ ∀nﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء.
I3
)
-4أﺳﺘﻨﺘﺞ + 2x 2 − 2x e x dx ﺗﻤﺮﻳﻦ4 1 -1ﺑﻴﻦ أن ≤ 1 ≤ 1− x 1+ x
3
∫0 ( x 1
∀x ∈ \ +
x2 -2اﺳﺘﻨﺘﺞ ≤ ln (1 + x ) ≤ x 2
∀x ∈ \ +
x−
2 ∫0 ln (1 + x ) dx 1
-3اﺳﺘﻨﺘﺞ ﺗﺄﻃﻴﺮا ﻟـ
إﻟﻰ .0,1
ﺗﻤﺮﻳﻦ9 2 2x = − 2 -1ﺗﺤﻘﻖ أن x x +1 x x 2 +1
2
)
-2ﻧﻌﺘﺒﺮ ]. k ∈ [ 0;1
*\ ∈ ∀x
(
ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء أﺣﺴﺐ dx
2x ln x 2
)
x 2 +1
1
(
k
∫ = Ak
ﺣﺪد lim Ak x →0
ﺗﻤﺮﻳﻦ10 2
t −t +1
1 1 -1أ -ﺗﺄآﺪ أن = − 2 t t 2 +1 t t +1
)
t2 − t +1
ب -أﺣﺴﺐ dt
)
3
(
t t2 +1
(
∫1
)
-2أﺣﺴﺐ + 2x + 1 ln ( x + 1)dx
2
∫0 ( 3x 1
ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء
ﺗﻤﺮﻳﻦ11 1 x -1ﺗﺄآﺪ أن − 2 x x +1
-2أﺣﺴﺐ dx
=
)
x ln x 2
)
+1
2
1
(x
2
(
x x +1
1
*\ ∈ ∀x
∫ = ) I (αﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء ﺣﻴﺚ [α ∈ ]0;1
α
-3أﺣﺴﺐ ) lim I (α
α → 0+
ﺗﻤﺮﻳﻦ12
4
3
π
∫02 x sin
π
1 I0 = ∫ 3 dx 0 cos x
;I n = ∫
n sin x ) ( 3
π
0
cos x
; I 3 و اﺳﺘﻨﺘﺞI1 أﺣﺴﺐ-1
I5 π
. n ﺑﺪﻻﻟﺔI n + 2 − I n و اﺳﺘﻨﺘﺞ
π 0; 3 ﻋﻠﻰ
x→
n ∈ `* ﻧﻌﺘﺒﺮ
dx و
∫03 ( sin x )
n
cos xdx أﺣﺴﺐ-2
1 x π داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔx → ln tg + ﺑﻴﻦ أن اﻟﺪاﻟﺔ- أ-3 cos x 2 4 I 4 ; I 2 ﺛﻢI 0 اﺳﺘﻨﺘﺞ-ب
5