✏️ Integral Hisob — Masalalar
1-bo'lim: Asosiy integrallar (5 ta masala)
Masala 1.1 ⭐⭐
Integralni hisoblang:
∫(3x2−4x+5)dx
Yechim
Har bir hadni alohida integrallaymiz:
∫3x2dx−∫4xdx+∫5dx
=3⋅3x3−4⋅2x2+5x+C
=x3−2x2+5x+C
Javob: x3−2x2+5x+C
Masala 1.2 ⭐⭐
∫x1dx
Yechim
∫x−1/2dx=1/2x1/2+C=2x+C
Javob: 2x+C
Masala 1.3 ⭐⭐
∫(ex+sinx)dx
Yechim
∫exdx+∫sinxdx=ex−cosx+C
Javob: ex−cosx+C
Masala 1.4 ⭐⭐
∫x3dx
Yechim
3∫x1dx=3ln∣x∣+C
Javob: 3ln∣x∣+C
Masala 1.5 ⭐⭐⭐
∫sec2x+1+x21dx
Yechim
∫sec2xdx+∫1+x21dx=tanx+arctanx+C
Javob: tanx+arctanx+C
2-bo'lim: Almashtiruv usuli (6 ta masala)
Masala 2.1 ⭐⭐
∫2x⋅(x2+1)5dx
Yechim
u=x2+1, du=2xdx
∫u5du=6u6+C=6(x2+1)6+C
Javob: 6(x2+1)6+C
Masala 2.2 ⭐⭐
∫cos(3x)dx
Yechim
u=3x, du=3dx, dx=3du
∫cosu⋅3du=31sinu+C=3sin(3x)+C
Javob: 3sin(3x)+C
Masala 2.3 ⭐⭐⭐
∫xexdx
Yechim
u=x, du=2x1dx, xdx=2du
∫eu⋅2du=2eu+C=2ex+C
Javob: 2ex+C
Masala 2.4 ⭐⭐⭐
∫x2+4xdx
Yechim
u=x2+4, du=2xdx, xdx=2du
∫u1⋅2du=21ln∣u∣+C=2ln(x2+4)+C
Javob: 2ln(x2+4)+C
Masala 2.5 ⭐⭐⭐
∫tanxdx
Yechim
∫cosxsinxdx
u=cosx, du=−sinxdx
−∫udu=−ln∣u∣+C=−ln∣cosx∣+C=ln∣secx∣+C
Javob: ln∣secx∣+C
Masala 2.6 ⭐⭐⭐
∫x⋅e−x2dx
Yechim
u=−x2, du=−2xdx, xdx=−2du
−21∫eudu=−2eu+C=−2e−x2+C
Javob: −2e−x2+C
3-bo'lim: Qismlab integrallash (5 ta masala)
Masala 3.1 ⭐⭐⭐
∫x⋅cosxdx
Yechim
u=x, dv=cosxdx
du=dx, v=sinx
∫xcosxdx=xsinx−∫sinxdx=xsinx+cosx+C
Javob: xsinx+cosx+C
Masala 3.2 ⭐⭐⭐
∫x2⋅exdx
Yechim
Ikki marta qismlab:
- u=x2, dv=exdx → du=2xdx, v=ex
x2ex−2∫xexdx
- ∫xexdx: u=x, dv=exdx
xex−ex
Natija:
x2ex−2(xex−ex)+C=ex(x2−2x+2)+C
Javob: ex(x2−2x+2)+C
Masala 3.3 ⭐⭐⭐
∫lnxdx
Yechim
u=lnx, dv=dx
du=xdx, v=x
xlnx−∫x⋅xdx=xlnx−x+C=x(lnx−1)+C
Javob: x(lnx−1)+C
Masala 3.4 ⭐⭐⭐⭐
∫exsinxdx
Yechim
I=∫exsinxdx
- u=sinx, dv=exdx
I=exsinx−∫excosxdx
- ∫excosxdx: u=cosx, dv=exdx
=excosx+∫exsinxdx=excosx+I
Qo'yamiz:
I=exsinx−excosx−I
2I=ex(sinx−cosx)
I=2ex(sinx−cosx)+C
Javob: 2ex(sinx−cosx)+C
Masala 3.5 ⭐⭐⭐⭐
∫x⋅arctanxdx
Yechim
u=arctanx, dv=xdx
du=1+x2dx, v=2x2
2x2arctanx−∫2(1+x2)x2dx
=2x2arctanx−21∫1+x2x2dx
=2x2arctanx−21∫(1−1+x21)dx
=2x2arctanx−2x+2arctanx+C
=2(x2+1)arctanx−x+C
Javob: 2(x2+1)arctanx−x+C
4-bo'lim: Aniq integrallar (5 ta masala)
Masala 4.1 ⭐⭐
∫02(3x2+1)dx
Yechim
[x3+x]02=(8+2)−(0+0)=10
Javob: 10
Masala 4.2 ⭐⭐
∫0π/2cosxdx
Yechim
[sinx]0π/2=sin(π/2)−sin(0)=1−0=1
Javob: 1
Masala 4.3 ⭐⭐⭐
∫1ex1dx
Yechim
[lnx]1e=lne−ln1=1−0=1
Javob: 1
Masala 4.4 ⭐⭐⭐
∫01x⋅ex2dx
Yechim
u=x2, du=2xdx
Chegaralar: x=0→u=0, x=1→u=1
21∫01eudu=21[eu]01=2e−1
Javob: 2e−1≈0.859
Masala 4.5 ⭐⭐⭐⭐
∫0πx⋅sinxdx
Yechim
Qismlab: u=x, dv=sinxdx
[−xcosx]0π+∫0πcosxdx
=(−π⋅(−1)−0)+[sinx]0π
=π+(0−0)=π
Javob: π
5-bo'lim: Yuzalar (4 ta masala)
Masala 5.1 ⭐⭐⭐
y=x2 va y=4 orasidagi yuza.
Yechim
Kesishish nuqtalari: x2=4 → x=±2
A=∫−22(4−x2)dx=[4x−3x3]−22
=(8−38)−(−8+38)=16−316=332
Javob: 332≈10.67 birlik kvadrat
Masala 5.2 ⭐⭐⭐
y=sinx va x-o'qi orasidagi yuza, 0 dan π gacha.
Yechim
A=∫0π∣sinx∣dx=∫0πsinxdx
=[−cosx]0π=−(−1)−(−1)=2
Javob: 2 birlik kvadrat
Masala 5.3 ⭐⭐⭐⭐
y=x va y=x2 orasidagi yuza.
Yechim
Kesishish: x=x2 → x(x−1)=0 → x=0,1
[0,1] oralig'ida x≥x2
A=∫01(x−x2)dx=[2x2−3x3]01=21−31=61
Javob: 61≈0.167 birlik kvadrat
Masala 5.4 ⭐⭐⭐⭐
y=ex, y=1, x=0, x=1 orasidagi yuza.
Yechim
[0,1] oralig'ida ex≥1
A=∫01(ex−1)dx=[ex−x]01=(e−1)−(1−0)=e−2
Javob: e−2≈0.718 birlik kvadrat
6-bo'lim: Hajmlar (4 ta masala)
Masala 6.1 ⭐⭐⭐
y=x, 0≤x≤4 ni x-o'qi atrofida aylantiring. Hajmni toping.
Yechim
V=π∫04(x)2dx=π∫04xdx=π[2x2]04=π⋅8=8π
Javob: 8π≈25.13 birlik kub
Masala 6.2 ⭐⭐⭐
y=x2, 0≤x≤2 ni x-o'qi atrofida aylantiring.
Yechim
V=π∫02(x2)2dx=π∫02x4dx=π[5x5]02=532π
Javob: 532π≈20.11 birlik kub
Masala 6.3 ⭐⭐⭐⭐
Shar hajmi (R radiusli).
Yechim
Shar: y=R2−x2 ni x-o'qi atrofida aylantirish.
V=π∫−RR(R2−x2)dx=π[R2x−3x3]−RR
=π[(R3−3R3)−(−R3+3R3)]
=π⋅2(R3−3R3)=34πR3
Javob: V=34πR3
Masala 6.4 ⭐⭐⭐⭐
Konus hajmi (balandlik h, radius R).
Yechim
To'g'ri chiziq: y=hRx, 0≤x≤h
V=π∫0h(hRx)2dx=h2πR2∫0hx2dx
=h2πR2⋅3h3=3πR2h
Javob: V=31πR2h
7-bo'lim: Fizik masalalar (6 ta masala)
Masala 7.1 ⭐⭐⭐
Robot tezligi: v(t)=3t2−2t m/s. t=0 dan t=3s gacha bosib o'tgan masofa.
Yechim
s=∫03∣v(t)∣dt
v(t)=t(3t−2)=0 → t=0 va t=2/3
[0,2/3]: v≤0, [2/3,3]: v≥0
s=−∫02/3(3t2−2t)dt+∫2/33(3t2−2t)dt
=−[t3−t2]02/3+[t3−t2]2/33
=−(278−94)+(27−9−278+94)
=274+18−274=18 m
Javob: 18 m
Masala 7.2 ⭐⭐⭐
Raketa tezlanishi: a(t)=20−2t m/s². Boshlang'ich tezlik v0=0. t=5s dagi tezlik va siljish.
Yechim
Tezlik:
v(t)=∫a(t)dt=20t−t2+C
v(0)=0 → C=0
v(5)=100−25=75 m/s
Siljish:
s=∫05(20t−t2)dt=[10t2−3t3]05
=250−3125=3625≈208.3 m
Javob: v=75 m/s, s≈208.3 m
Masala 7.3 ⭐⭐⭐
Prujina kuchi: F(x)=kx, k=100 N/m. Prujinani x=0 dan x=0.2 m ga cho'zish uchun ish.
Yechim
W=∫00.2F(x)dx=∫00.2100xdx
=[50x2]00.2=50⋅0.04=2 J
Javob: W=2 J
Masala 7.4 ⭐⭐⭐⭐
Quadcopter quvvat sarfi: P(t)=200+50sin(t) W. 10 sekund davomida sarflangan energiya.
Yechim
E=∫010P(t)dt=∫010(200+50sint)dt
=[200t−50cost]010
=(2000−50cos10)−(0−50)
=2050−50cos10≈2050−50(−0.839)≈2092 J
Javob: E≈2092 J ≈2.1 kJ
Masala 7.5 ⭐⭐⭐⭐
Raketa yoqilg'i sarfi: m˙(t)=100e−0.1t kg/s. 10 sekund davomida sarflangan yoqilg'i.
Yechim
m=∫010100e−0.1tdt=[−1000e−0.1t]010
=−1000e−1+1000=1000(1−e−1)
≈1000(1−0.368)=632 kg
Javob: m≈632 kg
Masala 7.6 ⭐⭐⭐⭐
PID kontroller integral termi: e(t)=5e−0.5t, Ki=2. t=0 dan t=4s gacha integral qiymati.
Yechim
uI=Ki∫04e(t)dt=2∫045e−0.5tdt
=10∫04e−0.5tdt=10⋅[−2e−0.5t]04
=−20[e−2−1]=20(1−e−2)
≈20(1−0.135)=17.3
Javob: uI≈17.3
8-bo'lim: Noto'g'ri integrallar (3 ta masala)
Masala 8.1 ⭐⭐⭐
∫1∞x21dx
Yechim
limb→∞∫1bx−2dx=limb→∞[−x−1]1b
=limb→∞(−b1+1)=0+1=1
Javob: Integral yaqinlashadi, =1
Masala 8.2 ⭐⭐⭐
∫0∞e−xdx
Yechim
limb→∞∫0be−xdx=limb→∞[−e−x]0b
=limb→∞(−e−b+1)=0+1=1
Javob: 1
Masala 8.3 ⭐⭐⭐⭐
∫01x1dx
Yechim
x=0 da uzilish mavjud.
limϵ→0+∫ϵ1x−1/2dx=limϵ→0+[2x]ϵ1
=limϵ→0+(2−2ϵ)=2−0=2
Javob: 2
9-bo'lim: Raqamli integrallash (2 ta masala)
Masala 9.1 ⭐⭐⭐
Trapetsiya qoidasi bilan hisoblang (n=4):
∫02e−x2dx
Yechim
h=42−0=0.5
xi: 0, 0.5, 1.0, 1.5, 2.0
f(xi): 1, 0.779, 0.368, 0.105, 0.018
I≈20.5[f(0)+2f(0.5)+2f(1)+2f(1.5)+f(2)]
=0.25[1+1.558+0.736+0.210+0.018]
=0.25×3.522=0.881
Javob: ≈0.881 (aniq qiymat: ≈0.882)
Masala 9.2 ⭐⭐⭐⭐
Simpson qoidasi bilan hisoblang (n=4):
∫01sin(πx)dx
Yechim
h=0.25
xi: 0, 0.25, 0.5, 0.75, 1.0
f(xi): 0, 0.707, 1, 0.707, 0
Simpson:
I≈3h[f0+4f1+2f2+4f3+f4]
=30.25[0+4(0.707)+2(1)+4(0.707)+0]
=30.25[5.656+2]=30.25×7.656=0.638
Aniq qiymat: π2=0.637
Javob: ≈0.638
Qiyinlik darajasi bo'yicha
| Daraja | Masalalar soni |
|---|
| ⭐⭐ Oson | 9 |
| ⭐⭐⭐ O'rta | 16 |
| ⭐⭐⭐⭐ Murakkab | 10 |
| Jami | 35 |