∫ (from 0 to 2) of (1+x^2)^2 wrt x eching

$$\int _ { 0 } ^ { 2 } ( 1 + x ^ { 2 } ) ^ { 2 } d x$$

Baholash

$$\frac{206}{15}\approx 13,733333333$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(1+x^{2}\right)^{2} kengaytirilishi uchun ishlating.

$$\int _{0}^{2}1+2x^{2}+\left(x^{2}\right)^{2}\mathrm{d}x$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\int _{0}^{2}1+2x^{2}+x^{4}\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int 1+2x^{2}+x^{4}\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 1\mathrm{d}x+\int 2x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int 1\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 1 integralini toping.

$$x+2\int x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{2}\mathrm{d}x integralni \frac{x^{3}}{3} bilan almashtiring. 2 ni \frac{x^{3}}{3} marotabaga ko'paytirish.

$$x+\frac{2x^{3}}{3}+\int x^{4}\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{4}\mathrm{d}x integralni \frac{x^{5}}{5} bilan almashtiring.

$$x+\frac{2x^{3}}{3}+\frac{x^{5}}{5}$$

Qisqartirish.

$$\frac{x^{5}}{5}+\frac{2x^{3}}{3}+x$$

Xos integral bu integral hisoblashning yuqori chegarasida hisoblangan ifodaning boshlangʻich holatidan chiqarib tashlagan holda integral hisoblashning quyi chegarasida hisoblangan ifodaning boshlangʻich holatidir.

$$\frac{2^{5}}{5}+\frac{2}{3}\times 2^{3}+2-\left(\frac{0^{5}}{5}+\frac{2}{3}\times 0^{3}+0\right)$$

Qisqartirish.

$$\frac{206}{15}$$