∫ (from 4 to 9) of x^3+2x+1 eching

$$\int _ { 4 } ^ { 9 } x ^ { 3 } + 2 x + 1$$

Baholash

$$\frac{6585}{4}=1646,25$$

Yechim qadamlari

Avval noaniq integralni baholang.

$$\int x^{3}+2x+1\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x^{3}\mathrm{d}x+\int 2x\mathrm{d}x+\int 1\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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.

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

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

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

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

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

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

$$\frac{9^{4}}{4}+9^{2}+9-\left(\frac{4^{4}}{4}+4^{2}+4\right)$$

Qisqartirish.

$$\frac{6585}{4}$$