∫ (from 4 to 9) of (sqrt{x}+1)sqrt{x wrt x eching

$$\int _ { 4 } ^ { 9 } ( \sqrt { x } + 1 ) \sqrt { x } d x$$

Baholash

$$\frac{271}{6}\approx 45,166666667$$

Yechim qadamlari

\sqrt{x}+1 ga \sqrt{x} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int _{4}^{9}\left(\sqrt{x}\right)^{2}+\sqrt{x}\mathrm{d}x$$

2 daraja ko‘rsatkichini \sqrt{x} ga hisoblang va x ni qiymatni oling.

$$\int _{4}^{9}x+\sqrt{x}\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int x+\sqrt{x}\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x\mathrm{d}x+\int \sqrt{x}\mathrm{d}x$$

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.

$$\frac{x^{2}}{2}+\int \sqrt{x}\mathrm{d}x$$

\sqrt{x} ni x^{\frac{1}{2}} sifatida qaytadan yozish. k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{\frac{1}{2}}\mathrm{d}x integralni \frac{x^{\frac{3}{2}}}{\frac{3}{2}} bilan almashtiring. Qisqartirish.

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

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

$$\frac{9^{2}}{2}+\frac{2}{3}\times 9^{\frac{3}{2}}-\left(\frac{4^{2}}{2}+\frac{2}{3}\times 4^{\frac{3}{2}}\right)$$

Qisqartirish.

$$\frac{271}{6}$$