∫ (from - 1 to 4) of x^2-6x+4 wrt x eching

$$\int _ { - 1 } ^ { 4 } x ^ { 2 } - 6 x + 4 d x$$

Baholash

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

Yechim qadamlari

Avval noaniq integralni baholang.

$$\int x^{2}-6x+4\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 4\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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.

$$\frac{x^{3}}{3}-6\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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. -6 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

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

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 4 integralini toping.

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

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

$$\frac{4^{3}}{3}-3\times 4^{2}+4\times 4-\left(\frac{\left(-1\right)^{3}}{3}-3\left(-1\right)^{2}+4\left(-1\right)\right)$$

Qisqartirish.

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