∫ (from 0 to 1) of (1-8v^3+16v^7)dv eching

$$\int _ { 0 } ^ { 1 } ( 1 - 8 v ^ { 3 } + 16 v ^ { 7 } ) d v$$

Baholash

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Yechim qadamlari

Avval noaniq integralni baholang.

$$\int 1-8v^{3}+16v^{7}\mathrm{d}v$$

Summani muddatma-muddat integratsiya qiling.

$$\int 1\mathrm{d}v+\int -8v^{3}\mathrm{d}v+\int 16v^{7}\mathrm{d}v$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int 1\mathrm{d}v-8\int v^{3}\mathrm{d}v+16\int v^{7}\mathrm{d}v$$

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

$$v-8\int v^{3}\mathrm{d}v+16\int v^{7}\mathrm{d}v$$

k\neq -1 uchun integral \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} boʻlgani uchun, \int v^{3}\mathrm{d}v integralni \frac{v^{4}}{4} bilan almashtiring. -8 ni \frac{v^{4}}{4} marotabaga ko'paytirish.

$$v-2v^{4}+16\int v^{7}\mathrm{d}v$$

k\neq -1 uchun integral \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} boʻlgani uchun, \int v^{7}\mathrm{d}v integralni \frac{v^{8}}{8} bilan almashtiring. 16 ni \frac{v^{8}}{8} marotabaga ko'paytirish.

$$v-2v^{4}+2v^{8}$$

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

$$1-2\times 1^{4}+2\times 1^{8}-\left(0-2\times 0^{4}+2\times 0^{8}\right)$$

Qisqartirish.

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