∫ (from 0 to 1) of {left(1-sqrt{x}right)}^2 wrt x eching

$$\int_{ 0 }^{ 1 } { \left(1- \sqrt{ x } \right) }^{ 2 } d x$$

Baholash

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

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(1-\sqrt{x}\right)^{2} kengaytirilishi uchun ishlating.

$$\int _{0}^{1}1-2\sqrt{x}+\left(\sqrt{x}\right)^{2}\mathrm{d}x$$

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

$$\int _{0}^{1}1-2\sqrt{x}+x\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int 1-2\sqrt{x}+x\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 1\mathrm{d}x+\int -2\sqrt{x}\mathrm{d}x+\int x\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int 1\mathrm{d}x-2\int \sqrt{x}\mathrm{d}x+\int x\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 1 integralini toping.

$$x-2\int \sqrt{x}\mathrm{d}x+\int x\mathrm{d}x$$

\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. -2 ni \frac{2x^{\frac{3}{2}}}{3} marotabaga ko'paytirish.

$$x-\frac{4x^{\frac{3}{2}}}{3}+\int x\mathrm{d}x$$

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.

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

Qisqartirish.

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

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

$$\frac{1^{2}}{2}-\frac{4}{3}\times 1^{\frac{3}{2}}+1-\left(\frac{0^{2}}{2}-\frac{4}{3}\times 0^{\frac{3}{2}}+0\right)$$

Qisqartirish.

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