∫ _{-frac{sqrt{6}}{2}-1}^sqrt{6/2-1}2x^2+2x-1 wrt x eching

$$\int_{ - \frac{ \sqrt{ 6 } }{ 2 } -1 }^{ \frac{ \sqrt{ 6 } }{ 2 } -1 } 2 { x }^{ 2 } +2x-1 d x$$

Baholash

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Yechim qadamlari

Avval noaniq integralni baholang.

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

Summani muddatma-muddat integratsiya qiling.

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

Har bir shartda konstantani qavsdan tashqariga oling.

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

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

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

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.

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

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

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

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

$$\frac{2}{3}\left(\frac{1}{2}\times 6^{\frac{1}{2}}-1\right)^{3}+\left(\frac{1}{2}\times 6^{\frac{1}{2}}-1\right)^{2}-\left(\frac{1}{2}\times 6^{\frac{1}{2}}-1\right)-\left(\frac{2}{3}\left(-\frac{1}{2}\times 6^{\frac{1}{2}}-1\right)^{3}+\left(-\frac{1}{2}\times 6^{\frac{1}{2}}-1\right)^{2}-\left(-\frac{1}{2}\times 6^{\frac{1}{2}}-1\right)\right)$$

Qisqartirish.

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