∫ (from - 5 to 1) of x^2-6x+5 eching

$$\int _ { - 5 } ^ { 1 } x ^ { 2 } - 6 x + 5$$

Baholash

$$144$$

Yechim qadamlari

Avval noaniq integralni baholang.

$$\int x^{2}-6x+5\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 5\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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.

$$\frac{x^{3}}{3}-6\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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.

$$\frac{x^{3}}{3}-3x^{2}+\int 5\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 5 integralini toping.

$$\frac{x^{3}}{3}-3x^{2}+5x$$

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

$$\frac{1^{3}}{3}-3\times 1^{2}+5\times 1-\left(\frac{\left(-5\right)^{3}}{3}-3\left(-5\right)^{2}+5\left(-5\right)\right)$$

Qisqartirish.

$$144$$