∫ (from 0 to 2) of (x^3+2x^2-3) wrt x= eching

$$\int _ { 0 } ^ { 2 } ( x ^ { 3 } + 2 x ^ { 2 } - 3 ) d x =$$

Baholash

$$\frac{10}{3}\approx 3,333333333$$

Yechim qadamlari

Avval noaniq integralni baholang.

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

Summani muddatma-muddat integratsiya qiling.

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

Har bir shartda konstantani qavsdan tashqariga oling.

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

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.

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

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

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

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, -3 integralini toping.

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

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

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

Qisqartirish.

$$\frac{10}{3}$$