∫ (from 0 to 1) of (2x+3)*(3x-5) wrt x eching

$$\int _ { 0 } ^ { 1 } ( 2 x + 3 ) \cdot ( 3 x - 5 ) d x$$

Baholash

$$-\frac{27}{2}=-13,5$$

Yechim qadamlari

2x+3 ifodaning har bir elementini 3x-5 ifodaning har bir elementiga ko‘paytirish orqali taqsimot qonuni xususiyatlarini qo‘llash mumkin.

$$\int _{0}^{1}6x^{2}-10x+9x-15\mathrm{d}x$$

-x ni olish uchun -10x va 9x ni birlashtirish.

$$\int _{0}^{1}6x^{2}-x-15\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int 6x^{2}-x-15\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 6x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int -15\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$6\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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

$$2x^{3}-\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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. -1 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

$$2x^{3}-\frac{x^{2}}{2}+\int -15\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, -15 integralini toping.

$$2x^{3}-\frac{x^{2}}{2}-15x$$

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

$$2\times 1^{3}-\frac{1^{2}}{2}-15-\left(2\times 0^{3}-\frac{0^{2}}{2}-15\times 0\right)$$

Qisqartirish.

$$-\frac{27}{2}$$