∫ (from 1 to 2) of 2x^2+3x-3 wrt x eching

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

Baholash

$$\frac{37}{6}\approx 6,166666667$$

Yechim qadamlari

Avval noaniq integralni baholang.

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

Summani muddatma-muddat integratsiya qiling.

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

Har bir shartda konstantani qavsdan tashqariga oling.

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

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.

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

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x\mathrm{d}x integralni \frac{x^{2}}{2} bilan almashtiring. 3 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

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

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

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

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

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

Qisqartirish.

$$\frac{37}{6}$$