∫ 1/4x^3-1/3x^2+1/2x wrt x eching

$$\int{ \frac{ 1 }{ 4 } { x }^{ 3 } - \frac{ 1 }{ 3 } { x }^{ 2 } + \frac{ 1 }{ 2 } x }d x$$

Baholash

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}+С$$

Yechim qadamlari

Summani muddatma-muddat integratsiya qiling.

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

Har bir shartda konstantani qavsdan tashqariga oling.

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

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{1}{4} ni \frac{x^{4}}{4} marotabaga ko'paytirish.

$$\frac{x^{4}}{16}-\frac{\int x^{2}\mathrm{d}x}{3}+\frac{\int x\mathrm{d}x}{2}$$

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

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{\int x\mathrm{d}x}{2}$$

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{1}{2} ni \frac{x^{2}}{2} marotabaga ko'paytirish.

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}$$

Агар F\left(x\right)f\left(x\right) ning dastlabki holati boʻlsa, u holatda f\left(x\right) ning barcha dastlabki holatlari toʻplami F\left(x\right)+C tarafidan belgilanadi. Shu sababli natijaga C\in \mathrm{R} integrallash konstantasini qoʻshing.

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}+С$$

x ga nisbatan hosilani topish

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