∫ (from 0 to 2) of (24+24x+06x^2)(x) wrt x eching

$$\int_{ 0 }^{ 2 } (24+24x+06 { x }^{ 2 } )(x) d x$$

Baholash

$$112$$

Yechim qadamlari

0 hosil qilish uchun 0 va 6 ni ko'paytirish.

$$\int _{0}^{2}\left(24+24x+0x^{2}\right)x\mathrm{d}x$$

Har qanday sonni nolga ko‘paytirsangiz, nol chiqadi.

$$\int _{0}^{2}\left(24+24x+0\right)x\mathrm{d}x$$

24 olish uchun 24 va 0'ni qo'shing.

$$\int _{0}^{2}\left(24+24x\right)x\mathrm{d}x$$

24+24x ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int _{0}^{2}24x+24x^{2}\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int 24x+24x^{2}\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 24x\mathrm{d}x+\int 24x^{2}\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$24\int x\mathrm{d}x+24\int x^{2}\mathrm{d}x$$

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. 24 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

$$12x^{2}+24\int x^{2}\mathrm{d}x$$

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

$$12x^{2}+8x^{3}$$

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

$$12\times 2^{2}+8\times 2^{3}-\left(12\times 0^{2}+8\times 0^{3}\right)$$

Qisqartirish.

$$112$$