Aniq integral

Bu nima va nega kerak

Boshlang'ich funksiya topish (noaniq integral) $F(x) + C$ ko'rinishidagi cheksiz ko'p funksiyalar oilasini beradi. Amaliy masalalarda esa ko'pincha aniq bir sonli natija zarur bo'ladi:

• O'zgaruvchan tezlik bilan harakatlanayotgan jism bosib o'tgan masofa ($s = \int_{t_1}^{t_2} v(t)\,dt$);
• Egri chiziqlar bilan chegaralangan notekis figuralar yuzasi;
• O'zgaruvchan kuch bajargan mexanik ish.

Bunday masalalarni yechish uchun soha cheksiz ko'p kichik qismlarga bo'linib, hosil bo'lgan yig'indining limiti olinadi. Bu tushuncha aniq integral deb ataladi.

Nyuton va Leybnits tomonidan kashf etilgan formula cheksiz yig'indilarni hisoblash o'rniga, shunchaki boshlang'ich funksiyaning oraliq chegaralaridagi qiymatlari ayirmasini olish kifoya ekanini ko'rsatdi.

Asosiy g'oya

1. Aniq integral va egri chiziqli trapetsiya

$[a; b]$ kesmada uzluksiz va $f(x) \ge 0$ funksiya berilgan bo'lsin. Yuqoridan $y = f(x)$ grafigi, pastdan $Ox$ o'qi, ikki yonidan $x = a$ va $x = b$ to'g'ri chiziqlari bilan chegaralangan shakl egri chiziqli trapetsiya deyiladi.

Kesmani $n$ ta bo'lakka bo'lib, to'g'ri to'rtburchaklar yuzalari yig'indisi limitini olsak, aniq integral hosil bo'ladi: $\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^n f(\xi_i)\,\Delta x_i$

Belgilashlar:

• $a$ — quyi chegara, $b$ — yuqori chegara;
• $f(x)$ — integral ostidagi funksiya, $dx$ — differensial.

Farqi: Noaniq integral — bu funksiyalar to'plami ($F(x) + C$), aniq integral esa — haqiqiy sondir.

2. Nyuton-Leybnits formulasi

Agar $F(x)$ funksiya $f(x)$ ning biror boshlang'ich funksiyasi bo'lsa ($F'(x) = f(x)$), u holda: $\int_a^b f(x)\,dx = \left. F(x) \right|_a^b = F(b) - F(a)$

Integrallash doimiysi $C$ aniq integralda qisqarib ketadi: $(F(b) + C) - (F(a) + C) = F(b) - F(a)$.

3. Boshlang'ich funksiyalar jadvali

$f(x)$	Boshlang'ich funksiya $F(x)$
$k$ (son)	$kx$
$x^n$ ($n \neq -1$)	$\dfrac{x^{n+1}}{n+1}$
$\dfrac{1}{x}$ ($x > 0$)	$\ln x$
$\dfrac{1}{\sqrt{x}}$	$2\sqrt{x}$
$e^{kx}$	$\dfrac{1}{k}e^{kx}$
$\sin(kx)$	$-\dfrac{1}{k}\cos(kx)$
$\cos(kx)$	$\dfrac{1}{k}\sin(kx)$
$\dfrac{1}{\cos^2 x}$	$\text{tg}\,x$

4. Aniq integralning xossalari

1. Chegaralar teng bo'lsa: $\int_a^a f(x)\,dx = 0$
2. Chegaralar almashsa: $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$
3. O'zgarmas son: $\int_a^b k f(x)\,dx = k \int_a^b f(x)\,dx$
4. Yig'indi/ayirma: $\int_a^b (f(x) \pm g(x))\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$
5. Oraliqni bo'lish: $\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$
6. Simmetrik oraliqda $[-a; a]$:
   • $f(x)$ toq bo'lsa: $\int_{-a}^a f(x)\,dx = 0$
   • $f(x)$ juft bo'lsa: $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$

Yechilgan misollar

1-misol. Ko'phadli integral

$\int_1^3 (3x^2 - 4x + 2)\,dx$ integralni hisoblang.

Yechish:

1. Har bir hadning boshlang'ich funksiyasini topamiz: $F(x) = x^3 - 2x^2 + 2x$
2. Nyuton-Leybnits formulasiga ko'ra chegaralarni qo'yamiz: $\left. (x^3 - 2x^2 + 2x) \right|_1^3 = F(3) - F(1)$
3. Qiymatlarni hisoblaymiz: $F(3) = 3^3 - 2 \cdot 3^2 + 2 \cdot 3 = 27 - 18 + 6 = 15$ $F(1) = 1^3 - 2 \cdot 1^2 + 2 \cdot 1 = 1 - 2 + 2 = 1$ $F(3) - F(1) = 15 - 1 = 14$

Javob: $14$.

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2-misol. Trigonometrik va ildizli integrallar

Hisoblang: a) $\int_0^{\frac{\pi}{3}} \sin(3x)\,dx$ b) $\int_1^4 \dfrac{1}{\sqrt{x}}\,dx$

Yechish: a) $\sin(3x)$ ning boshlang'ich funksiyasi $-\dfrac{1}{3}\cos(3x)$: $\left. \left(-\frac{1}{3}\cos(3x)\right) \right|_0^{\frac{\pi}{3}} = -\frac{1}{3}\cos\pi - \left(-\frac{1}{3}\cos 0\right) = -\frac{1}{3}(-1) + \frac{1}{3}(1) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

b) $\dfrac{1}{\sqrt{x}}$ ning boshlang'ich funksiyasi $2\sqrt{x}$: $\left. 2\sqrt{x} \right|_1^4 = 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2$

Javob: a) $\dfrac{2}{3}$; b) $2$.

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3-misol. Modulli funksiya integrali

$\int_0^3 |x - 2|\,dx$ integralni hisoblang.

Yechish: Modul ostidagi ifoda $x = 2$ da nolga aylanadi. Shuning uchun oraliqni ikkiga ajratamiz:

• $x \in [0; 2]$ da $x - 2 \le 0 \implies |x - 2| = 2 - x$;
• $x \in [2; 3]$ da $x - 2 \ge 0 \implies |x - 2| = x - 2$.

$\int_0^3 |x - 2|\,dx = \int_0^2 (2 - x)\,dx + \int_2^3 (x - 2)\,dx$

1. $\int_0^2 (2 - x)\,dx = \left. \left(2x - \dfrac{x^2}{2}\right) \right|_0^2 = 4 - 2 = 2$
2. $\int_2^3 (x - 2)\,dx = \left. \left(\dfrac{x^2}{2} - 2x\right) \right|_2^3 = \left(\dfrac{9}{2} - 6\right) - (2 - 4) = -1{,}5 - (-2) = 0{,}5$

Yig'indi: $2 + 0{,}5 = 2{,}5 = \dfrac{5}{2}$.

Javob: $2{,}5$ (yoki $\dfrac{5}{2}$).

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4-misol. Juft va toq funksiyalar qoidasi

$\int_{-2}^2 (5x^7 - 3x^3 + \sin x + 3x^2 - 1)\,dx$ integralni hisoblang.

Yechish: Chegaralar nolga nisbatan simmetrik: $[-2; 2]$.

1. $5x^7 - 3x^3 + \sin x$ toq funksiya bo'lgani sababli, simmetrik oraliqdagi integrali $0$ ga teng.
2. $3x^2 - 1$ juft funksiya bo'lib, uning integrali yarim oraliqdagi integralning ikkilanganiga teng: $\int_{-2}^2 (3x^2 - 1)\,dx = 2 \int_0^2 (3x^2 - 1)\,dx = 2 \left. (x^3 - x) \right|_0^2 = 2(8 - 2) = 12$

Natija: $0 + 12 = 12$.

Javob: $12$.

Mashq

Quyidagi aniq integrallarni hisoblang:

1. $\int_0^2 (3x^2 + 2x)\,dx$
2. $\int_1^3 (4x - 1)\,dx$
3. $\int_0^1 (6x^5 - 4x^3 + 1)\,dx$
4. $\int_1^4 (3\sqrt{x} + 2)\,dx$
5. $\int_0^{\frac{\pi}{4}} \dfrac{1}{\cos^2 x}\,dx$
6. $\int_0^{\frac{\pi}{2}} \sin(2x)\,dx$
7. $\int_{-3}^3 (x^5 - 2x + 4)\,dx$
8. $\int_0^4 |x - 1|\,dx$
9. $\int_1^e \dfrac{3}{x}\,dx$
10. Agar $\int_1^5 f(x)\,dx = 7$ va $\int_3^5 f(x)\,dx = 2$ bo'lsa, $\int_1^3 f(x)\,dx$ ni toping.

Javoblar va yechim yo'llari

1. $F(x) = x^3 + x^2$. $F(2) - F(0) = (8 + 4) - 0 = 12$. Javob: $12$.

2. $F(x) = 2x^2 - x$. $F(3) = 18 - 3 = 15$, $F(1) = 2 - 1 = 1$. Ayirma: $15 - 1 = 14$. Javob: $14$.

3. $F(x) = x^6 - x^4 + x$. $F(1) - F(0) = (1 - 1 + 1) - 0 = 1$. Javob: $1$.

4. $F(x) = 2x\sqrt{x} + 2x$. $F(4) = 16 + 8 = 24$, $F(1) = 2 + 2 = 4$. $24 - 4 = 20$. Javob: $20$.

5. $F(x) = \text{tg}\,x$. $F(\pi/4) - F(0) = 1 - 0 = 1$. Javob: $1$.

6. $F(x) = -\dfrac{1}{2}\cos(2x)$. $F(\pi/2) - F(0) = \dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 1$. Javob: $1$.

7. Toq qism integrali $0$. $\int_{-3}^3 4\,dx = \left. 4x \right|_{-3}^3 = 12 - (-12) = 24$. Javob: $24$.

8. $\int_0^1 (1 - x)\,dx + \int_1^4 (x - 1)\,dx = 0{,}5 + 4{,}5 = 5$. Javob: $5$.

9. $F(x) = 3\ln x$. $F(e) - F(1) = 3(1) - 3(0) = 3$. Javob: $3$.

10. Additivlik: $\int_1^3 f(x)\,dx = \int_1^5 f(x)\,dx - \int_3^5 f(x)\,dx = 7 - 2 = 5$. Javob: $5$.

Keng tarqalgan xatolar

1. Chegaralarni almashtirib qo'yish (quyi chegaradan yuqorini ayirish). ✗ $\int_1^3 2x\,dx = F(1) - F(3) = 1 - 9 = -8$ deb yozish. ✓ Nyuton-Leybnits formulasida qat'iy ravishda yuqori chegaradagi qiymatdan quyi chegaradagi qiymat ayiriladi: $F(b) - F(a) = 9 - 1 = 8$.

2. Murakkab argument koeffitsiyentini ($\frac{1}{k}$) unutish. ✗ $\int_0^{\pi} \cos(2x)\,dx = \left. \sin(2x) \right|_0^{\pi}$ deb yozish. ✓ Argument $kx + b$ ko'rinishida bo'lsa, boshlang'ich funksiya oldiga $\dfrac{1}{k}$ ko'paytuvchisi chiqadi: $\int_0^{\pi} \cos(2x)\,dx = \left. \dfrac{1}{2}\sin(2x) \right|_0^{\pi} = 0$.

3. Modulli funksiyani oraliqlarga ajratmasdan integrallash. ✗ $\int_0^2 |x - 1|\,dx = \left. \left|\dfrac{x^2}{2} - x\right| \right|_0^2 = |2 - 2| = 0$ deb xato qilish. ✓ Modul ichidagi ifodaning ishorasi o'zgaradigan nol nuqtasi ($x = 1$) orqali oraliq ikkiga ajratilib integrallanadi: to'g'ri natija $1$ ga teng.

4. Integrallash doimiysi $C$ ni aniq integralda ham qo'shib hisoblash. ✗ $\int_1^2 3x^2\,dx = (x^3 + C)|_1^2 = 8 + C - 1 - C$... deb $C$ ni izlash. ✓ Aniq integralda o'zgarmas $C$ doimo ayirmada qisqarib ketadi ($C - C = 0$), uni yozish shart emas.

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Keyingi qadam

Biz aniq integralni hisoblash qoidalari va Nyuton-Leybnits formulasi bilan tanishdik. Keyingi darsda aniq integralning eng asosiy geometrik tatbiqi — tekislikdagi murakkab shakllar va egri chiziqlar bilan chegaralangan sohalar yuzasini hisoblashni o'rganamiz.