∫ (from 0 to 1) of 4x(1+x)^3 wrt x eching

$$\int _ { 0 } ^ { 1 } 4 x ( 1 + x ) ^ { 3 } d x$$

Baholash

$$\frac{49}{5}=9,8$$

Yechim qadamlari

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

$$\int _{0}^{1}4x\left(1+3x+3x^{2}+x^{3}\right)\mathrm{d}x$$

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

$$\int _{0}^{1}4x+12x^{2}+12x^{3}+4x^{4}\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int 4x+12x^{2}+12x^{3}+4x^{4}\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 4x\mathrm{d}x+\int 12x^{2}\mathrm{d}x+\int 12x^{3}\mathrm{d}x+\int 4x^{4}\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$4\int x\mathrm{d}x+12\int x^{2}\mathrm{d}x+12\int x^{3}\mathrm{d}x+4\int x^{4}\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x\mathrm{d}x integralni \frac{x^{2}}{2} bilan almashtiring. 4 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

$$2x^{2}+12\int x^{2}\mathrm{d}x+12\int x^{3}\mathrm{d}x+4\int x^{4}\mathrm{d}x$$

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. 12 ni \frac{x^{3}}{3} marotabaga ko'paytirish.

$$2x^{2}+4x^{3}+12\int x^{3}\mathrm{d}x+4\int x^{4}\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{3}\mathrm{d}x integralni \frac{x^{4}}{4} bilan almashtiring. 12 ni \frac{x^{4}}{4} marotabaga ko'paytirish.

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

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. 4 ni \frac{x^{5}}{5} marotabaga ko'paytirish.

$$2x^{2}+4x^{3}+3x^{4}+\frac{4x^{5}}{5}$$

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

$$2\times 1^{2}+4\times 1^{3}+3\times 1^{4}+\frac{4}{5}\times 1^{5}-\left(2\times 0^{2}+4\times 0^{3}+3\times 0^{4}+\frac{4}{5}\times 0^{5}\right)$$

Qisqartirish.

$$\frac{49}{5}$$