∫ (from 3 to 4) of -x^2+7x-12 eching

$$\int _ { 3 } ^ { 4 } - x ^ { 2 } + 7 x - 12$$

Baholash

$$\frac{1}{6}\approx 0,166666667$$

Yechim qadamlari

Avval noaniq integralni baholang.

$$\int -x^{2}+7x-12\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int -x^{2}\mathrm{d}x+\int 7x\mathrm{d}x+\int -12\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$-\int x^{2}\mathrm{d}x+7\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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. -1 ni \frac{x^{3}}{3} marotabaga ko'paytirish.

$$-\frac{x^{3}}{3}+7\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$-\frac{x^{3}}{3}+\frac{7x^{2}}{2}+\int -12\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, -12 integralini toping.

$$-\frac{x^{3}}{3}+\frac{7x^{2}}{2}-12x$$

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

$$-\frac{4^{3}}{3}+\frac{7}{2}\times 4^{2}-12\times 4-\left(-\frac{3^{3}}{3}+\frac{7}{2}\times 3^{2}-12\times 3\right)$$

Qisqartirish.

$$\frac{1}{6}$$