Mavzular

⭐ — imtihonda eng ko'p ishlatiladigan formulalar. · 🟢 — masalani tezlashtiradi · 🟡 — ilg'or daraja · 🔵 — chuqur o'rganish (olimpiada darajasi)

Aniq integral va yuza

  • Egri chiziq ostidagi yuza AA uchun dAdx=y\dfrac{dA}{dx} = y. Shuning uchun yuza integrallash orqali topiladi.
  • Yuza — ingichka to'g'ri to'rtburchaklar yig'indisining limiti: A=lim⁡δx→0∑y δx=∫aby dxA = \lim_{\delta x \to 0} \sum y\,\delta x = \int_a^b y\,dx
  • ⭐ Aniq integral: ∫abf(x) dx=[F(x)]ab=F(b)−F(a)\displaystyle\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a), bu yerda F′(x)=f(x)F'(x) = f(x).
  • Chegaralarni almashtirsangiz, ishorasi o'zgaradi: ∫baf(x) dx=−∫abf(x) dx\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx
  • ⭐ Egri chiziq va xx o'qi orasidagi yuza: A=∫aby dx\displaystyle A = \int_a^b y\,dx
  • ⭐ xx o'qidan pastdagi yuza uchun integral manfiy chiqadi. Soha o'qning ikki tomonida bo'lsa, integralni ildizlar bo'yicha bo'lib hisoblang: A=∣∫apf(x) dx∣+∣∫pbf(x) dx∣(f(p)=0)A = \left|\int_a^p f(x)\,dx\right| + \left|\int_p^b f(x)\,dx\right| \qquad (f(p) = 0)
  • ⭐ Ikki egri chiziq orasidagi yuza (yuqoridagidan pastdagini ayirasiz): A=∫ab(f(x)−g(x)) dxA = \int_a^b \big(f(x) - g(x)\big)\,dx Chegaralar aa, bb — egri chiziqlar kesishgan nuqtalar: f(x)=g(x)f(x) = g(x) ni yechasiz.
  • Egri chiziq va yy o'qi orasidagi yuza: A=∫pqx dy\displaystyle A = \int_p^q x\,dy

Xosmas integrallar (improper)

  • Chegara cheksiz bo'lsa, limit orqali hisoblanadi: ∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^\infty f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx
  • Funksiya chegarada aniqlanmagan bo'lsa ham limit olinadi: ∫011x dx=lim⁡a→0∫a1x−12 dx=lim⁡a→0(2−2a)=2\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{a \to 0}\int_a^1 x^{-\frac{1}{2}}\,dx = \lim_{a \to 0}\big(2 - 2\sqrt{a}\big) = 2

🟢 Asosiy xossalar

∫abf(x) dx+∫bcf(x) dx=∫acf(x) dx∫aaf(x) dx=0\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx \qquad \int_a^a f(x)\,dx = 0

🟢 Simmetrik chegaralar

∫−aaf(x) dx=0    (f toq)∫−aaf(x) dx=2∫0af(x) dx    (f juft)\int_{-a}^{a} f(x)\,dx = 0 \;\;(f \text{ toq}) \qquad \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx \;\;(f \text{ juft})

Misol: ∫−22(x3+5x) dx=0\displaystyle\int_{-2}^{2} (x^3 + 5x)\,dx = 0 — hisoblamasdan.

🟡 Funksiyaning o'rtacha qiymati

fˉ=1b−a∫abf(x) dx\bar{f} = \frac{1}{b - a}\int_a^b f(x)\,dx

🔵 Integral — yig'indining limiti

∫abf(x) dx=lim⁡δx→0∑f(xi) δx\int_a^b f(x)\,dx = \lim_{\delta x \to 0}\sum f(x_i)\,\delta x

🟢 Parabolik segment yuzi

y=a(x−α)(x−β)y = a(x - \alpha)(x - \beta) va xx o'qi orasidagi yuza:

S=∣a∣ (β−α)36S = \frac{|a|\,(\beta - \alpha)^3}{6}

Shu formula parabola y=ax2+…y = ax^2 + \dots va istalgan to'g'ri chiziq orasidagi yuza uchun ham ishlaydi: α\alpha, β\beta — kesishish nuqtalarining xx koordinatalari. Misol: y=x2y = x^2 va y=4y = 4 orasida: α=−2\alpha = -2, β=2\beta = 2, S=1⋅436=323S = \dfrac{1 \cdot 4^3}{6} = \dfrac{32}{3} ✓

🟢 Birlik kvadratdagi yuzalar

∫01xn dx=1n+1\int_0^1 x^n\,dx = \frac{1}{n + 1}

Masalan, y=x2y = x^2 birlik kvadratni 13\frac{1}{3} va 23\frac{2}{3} ga bo'ladi.

🟢 Darajali funksiyalar uchun

∫1∞dxxn=1n−1    (n>1)∫01dxxn=11−n    (n<1)\int_1^\infty \frac{dx}{x^n} = \frac{1}{n - 1} \;\;(n > 1) \qquad \int_0^1 \frac{dx}{x^n} = \frac{1}{1 - n} \;\;(n < 1)

n≤1n \le 1 bo'lsa birinchisi, n≥1n \ge 1 bo'lsa ikkinchisi uzoqlashadi (cheksiz).

🔵 Egri chiziq uzunligi

L=∫ab1+(dydx)2 dxL = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx

Mashq qiling

15 ta savol, vaqt bilan, izohlari bilan.

Testga oʻtish
Yordam