Hosila va integral masalalari

Bu mavzuda DTM nimani so'raydi

Hosila va integral DTM testlarida har yili 2–4 ta savol bilan qatnashadi:

1. Hosilaning geometrik va fizik ma'nosi: urinmaning burchak koeffitsiyenti $k = f'(x_0)$, urinma tenglamasi, harakat tezligi $v(t) = s'(t)$ va tezlanishi $a(t) = v'(t)$.
2. Funksiyani tekshirish: o'sish va kamayish oraliqlari, maksimum va minimum nuqtalari.
3. Kesmada eng katta va eng kichik qiymat: statsionar nuqtalar va kesma chegaralaridagi qiymatlarni taqqoslash.
4. Aniq integral va yuza: Nyuton-Leybnits formulasi, egri chiziqlar bilan chegaralangan figura yuzi.

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Asosiy formulalar va usullar

1. Urinma va hosilaning qo'llanishi

• Burchak koeffitsiyenti: $k = \operatorname{tg}\alpha = f'(x_0)$
• Urinma tenglamasi: $y = f(x_0) + f'(x_0)(x - x_0)$
• Harakat tezlanishi: $a(t) = s''(t)$

2. O'sish, kamayish va ekstremum

• $f'(x) > 0$ da funksiya o'sadi, $f'(x) < 0$ da kamayadi.
• $f'(x_0) = 0$ bo'lib, ishora $+$ dan $-$ ga o'tsa — maksimum; $-$ dan $+$ ga o'tsa — minimum nuqta.

3. Integral va yuza

• $\int_a^b f(x)\,dx = F(b) - F(a)$
• Parabola segmenti yuzi (Arximed qoidasi): $S = \dfrac{2}{3} \cdot \text{asos} \cdot \text{balandlik}$

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Yechilgan misollar

1-misol

$f(x) = x^3 - 3x^2 + 5$ funksiyaning $[-1; 3]$ kesmadagi eng katta va eng kichik qiymatlari yig'indisini toping.

A) $4$
B) $5$
C) $6$
D) $7$

Yechimi:

1. $f'(x) = 3x^2 - 6x = 3x(x - 2) = 0 \implies x_1 = 0, x_2 = 2 \in [-1; 3]$.
2. Qiymatlarni hisoblaymiz:
   • $f(-1) = -1 - 3 + 5 = 1$
   • $f(0) = 5$
   • $f(2) = 8 - 12 + 5 = 1$
   • $f(3) = 27 - 27 + 5 = 5$
3. Eng katta qiymat $5$, eng kichik qiymat $1$. Yig'indi: $5 + 1 = 6$.

To'g'ri javob: C.

Tezroq yo'l: Ekstremumlar $f(0) = 5$ va $f(2) = 1$. Chegaralar ham $1$ va $5$. Yig'indi darhol $5 + 1 = 6$.

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2-misol

$y = x^2 - 4x + 3$ parabolaga uning abssissasi $x_0 = 3$ bo'lgan nuqtasida o'tkazilgan urinmaning burchak koeffitsiyentini toping.

A) $1$
B) $2$
C) $3$
D) $-2$

Yechimi:
$k = y'(x_0)$.
$y' = 2x - 4 \implies y'(3) = 2(3) - 4 = 2$.

To'g'ri javob: B.

Tezroq yo'l: $y' = 2x - 4$. $x = 3$ da og'zaki $6 - 4 = 2$.

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3-misol

$\int_0^2 (3x^2 - 2x + 1)\,dx$ aniq integralni hisoblang.

A) $4$
B) $5$
C) $6$
D) $8$

Yechimi:
Boshlang'ich funksiya: $F(x) = x^3 - x^2 + x$.
$\int_0^2 (3x^2 - 2x + 1)\,dx = [x^3 - x^2 + x]_0^2 = (8 - 4 + 2) - 0 = 6$

To'g'ri javob: C.

Tezroq yo'l: Hadma-had: $[x^3]_0^2 = 8$, $[-x^2]_0^2 = -4$, $[x]_0^2 = 2$. $8 - 4 + 2 = 6$.

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4-misol

$y = 4 - x^2$ parabola va $y = 0$ to'g'ri chiziq bilan chegaralangan sohaning yuzini toping.

A) $\dfrac{16}{3}$
B) $8$
C) $\dfrac{32}{3}$
D) $12$

Yechimi:
Kesishish nuqtalari: $4 - x^2 = 0 \implies x = \pm 2$.
$S = \int_{-2}^2 (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^2 = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{32}{3}$

To'g'ri javob: C.

Tezroq yo'l: Arximed formulasi: $S = \dfrac{2}{3} \cdot \text{asos} \cdot \text{balandlik} = \dfrac{2}{3} \cdot 4 \cdot 4 = \dfrac{32}{3}$.

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5-misol

$f(x) = \dfrac{x^3}{3} - 4x$ funksiyaning kamayish oralig'ining uzunligini toping.

A) $2$
B) $4$
C) $6$
D) $8$

Yechimi:
$f'(x) \le 0 \implies x^2 - 4 \le 0 \implies x \in [-2; 2]$.
Oraliq uzunligi: $2 - (-2) = 4$.

To'g'ri javob: B.

Tezroq yo'l: $x^2 \le 4 \implies [-2; 2]$. Uzunligi: $2 - (-2) = 4$.

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Mashq

1. $f(x) = 2x^3 - 9x^2 + 12x - 1$ funksiyaning maksimum nuqtasini toping.

Yechimi va javobi

$f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2) = 0$. $x = 1$ nuqtada hosila $+$ dan $-$ ga o'tadi $\implies x = 1$ maksimum nuqta.

Javob: $1$

2. $y = x^2 - 2x + 5$ parabolaga uning $x_0 = 2$ nuqtasida o'tkazilgan urinmaning tenglamasini yozing.

Yechimi va javobi

$y(2) = 5$. $y' = 2x - 2 \implies k = y'(2) = 2$.
Urinma: $y - 5 = 2(x - 2) \implies y = 2x + 1$.

Javob: $y = 2x + 1$

3. Hisoblang: $\int_1^4 \dfrac{1}{\sqrt{x}}\,dx$.

Yechimi va javobi

$[2\sqrt{x}]_1^4 = 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2$.

Javob: $2$

4. $f(x) = 2\sin x + \cos 2x$ funksiyaning $x_0 = \dfrac{\pi}{6}$ nuqtadagi hosilasini hisoblang.

Yechimi va javobi

$f'(x) = 2\cos x - 2\sin 2x$.
$f'\left(\dfrac{\pi}{6}\right) = 2\cos\dfrac{\pi}{6} - 2\sin\dfrac{\pi}{3} = \sqrt{3} - \sqrt{3} = 0$.

Javob: $0$

5. $y = x^2$ va $y = 2x$ chiziqlar bilan chegaralangan sohaning yuzini hisoblang.

Yechimi va javobi

Kesishish nuqtalari: $x = 0, x = 2$.
$S = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$

Javob: $\dfrac{4}{3}$

6. $F(x)$ funksiya $f(x) = 3x^2 - 4x + 1$ ning boshlang'ichi va grafigi $M(1; 3)$ nuqtadan o'tadi. $F(x)$ ni toping.

Yechimi va javobi

$F(x) = x^3 - 2x^2 + x + C$. $F(1) = 1 - 2 + 1 + C = C = 3 \implies F(x) = x^3 - 2x^2 + x + 3$.

Javob: $F(x) = x^3 - 2x^2 + x + 3$

7. $y = x^3 - 3x$ funksiya grafigiga o'tkazilgan qaysi urinma $y = 9x - 2$ to'g'ri chiziqqa parallel bo'ladi? Urinish nuqtalarining abssissalarini toping.

Yechimi va javobi

$k = 9$. $y' = 3x^2 - 3 = 9 \implies 3x^2 = 12 \implies x = \pm 2$.

Javob: $\pm 2$

8. Hisoblang: $\int_0^{\pi/2} \cos x\,dx$.

Yechimi va javobi

$[\sin x]_0^{\pi/2} = \sin\dfrac{\pi}{2} - \sin 0 = 1$.

Javob: $1$

9. $f(x) = x + \dfrac{4}{x}$ funksiyaning $x \in [1; 4]$ kesmadagi eng kichik qiymatini toping.

Yechimi va javobi

$f'(x) = 1 - \dfrac{4}{x^2} = 0 \implies x = 2$. Qiymatlar: $f(1) = 5, f(2) = 4, f(4) = 5$. Eng kichigi: $4$.

Javob: $4$

10. Nuqtaning harakat qonuni: $s(t) = 2t^3 - 3t^2 + 4t - 1$ ($t$ sekundda, $s$ metrda). $t = 2\text{ s}$ dagi tezlanishni toping.

Yechimi va javobi

$v(t) = s'(t) = 6t^2 - 6t + 4$. Tezlanish $a(t) = v'(t) = 12t - 6$. $a(2) = 12(2) - 6 = 18\text{ m/s}^2$.

Javob: $18\text{ m/s}^2$

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Keng tarqalgan tuzoqlar

1. Ekstremum nuqta bilan ekstremum qiymatni almashtirib qo'yish:
   "Maksimum nuqtasi" so'ralsa $x$ javob bo'ladi; "Maksimum qiymati" so'ralsa $y = f(x)$ qiymati topiladi.
2. Kesmada eng katta/kichik qiymatni topishda chegaralarni tekshirmaslik:
   Ekstremumlar har doim ham hosila nolga teng bo'lgan joyda emas, balki kesma uchlarida bo'lishi mumkin. $f(a)$ va $f(b)$ albatta hisoblanishi shart.
3. Urinma tenglamasidagi xato:
   Urinma formulasi $y = f(x_0) + f'(x_0)(x - x_0)$ ko'rinishida bo'lib, $f(x_0)$ qo'shilishi esdan chiqmasligi kerak.
4. Yuza hisoblashda manfiy ishora:
   Egri chiziqli shakl yuzi doimo musbat miqdordir. $Ox$ o'qidan pastdagi integrallarda minus ishorasini hisobga olib, yuza modul bilan yoki yuqori funksiyadan pastkisini ayirish orqali hisoblanadi.