∫ (from 1 to 2) of (x^2-1)^3x wrt x eching

$$\int _ { 1 } ^ { 2 } ( x ^ { 2 } - 1 ) ^ { 3 } x d x$$

Baholash

$$\frac{81}{8}=10,125$$

Yechim qadamlari

\left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} binom teoremasini \left(x^{2}-1\right)^{3} kengaytirilishi uchun ishlating.

$$\int _{1}^{2}\left(\left(x^{2}\right)^{3}-3\left(x^{2}\right)^{2}+3x^{2}-1\right)x\mathrm{d}x$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 3 ni ko‘paytirib, 6 ni oling.

$$\int _{1}^{2}\left(x^{6}-3\left(x^{2}\right)^{2}+3x^{2}-1\right)x\mathrm{d}x$$

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

$$\int _{1}^{2}\left(x^{6}-3x^{4}+3x^{2}-1\right)x\mathrm{d}x$$

x^{6}-3x^{4}+3x^{2}-1 ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int _{1}^{2}x^{7}-3x^{5}+3x^{3}-x\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int x^{7}-3x^{5}+3x^{3}-x\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x^{7}\mathrm{d}x+\int -3x^{5}\mathrm{d}x+\int 3x^{3}\mathrm{d}x+\int -x\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{7}\mathrm{d}x-3\int x^{5}\mathrm{d}x+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{7}\mathrm{d}x integralni \frac{x^{8}}{8} bilan almashtiring.

$$\frac{x^{8}}{8}-3\int x^{5}\mathrm{d}x+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{5}\mathrm{d}x integralni \frac{x^{6}}{6} bilan almashtiring. -3 ni \frac{x^{6}}{6} marotabaga ko'paytirish.

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{3}\mathrm{d}x integralni \frac{x^{4}}{4} bilan almashtiring. 3 ni \frac{x^{4}}{4} marotabaga ko'paytirish.

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+\frac{3x^{4}}{4}-\int x\mathrm{d}x$$

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

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+\frac{3x^{4}}{4}-\frac{x^{2}}{2}$$

Qisqartirish.

$$-\frac{x^{2}}{2}+\frac{3x^{4}}{4}-\frac{x^{6}}{2}+\frac{x^{8}}{8}$$

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

$$-\frac{2^{2}}{2}+\frac{3}{4}\times 2^{4}-\frac{2^{6}}{2}+\frac{2^{8}}{8}-\left(-\frac{1^{2}}{2}+\frac{3}{4}\times 1^{4}-\frac{1^{6}}{2}+\frac{1^{8}}{8}\right)$$

Qisqartirish.

$$\frac{81}{8}$$