∫ (from -2 to 3) of x((x^2)-(x+6)) wrt x eching

$$\int_{ -2 }^{ 3 } x(( { x }^{ 2 } )-(x+6)) d x$$

Baholash

$$-\frac{125}{12}\approx -10,416666667$$

Yechim qadamlari

x+6 teskarisini topish uchun har birining teskarisini toping.

$$\int _{-2}^{3}x\left(x^{2}-x-6\right)\mathrm{d}x$$

x ga x^{2}-x-6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int _{-2}^{3}x^{3}-x^{2}-6x\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int x^{3}-x^{2}-6x\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x^{3}\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -6x\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{3}\mathrm{d}x-\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{3}\mathrm{d}x integralni \frac{x^{4}}{4} bilan almashtiring.

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

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.

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

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.

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

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

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

Qisqartirish.

$$-\frac{125}{12}$$