∫ (from 0 to 6+3) of sqrt{x wrt x eching

$$\int_{ 0 }^{ 6+3 } \sqrt{ x } d x$$

Baholash

$$18$$

Yechim qadamlari

Avval noaniq integralni baholang.

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

\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.

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

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

$$\frac{2}{3}\left(6+3\right)^{\frac{3}{2}}-\frac{2}{3}\times 0^{\frac{3}{2}}$$

Qisqartirish.

$$18$$