∫ (4x-7)^2 wrt x eching

$$\int ( 4 x - 7 ) ^ { 2 } d x$$

Baholash

$$\frac{\left(4x-7\right)^{3}}{12}+С$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(4x-7\right)^{2} kengaytirilishi uchun ishlating.

$$\int 16x^{2}-56x+49\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 16x^{2}\mathrm{d}x+\int -56x\mathrm{d}x+\int 49\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$16\int x^{2}\mathrm{d}x-56\int x\mathrm{d}x+\int 49\mathrm{d}x$$

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

$$\frac{16x^{3}}{3}-56\int x\mathrm{d}x+\int 49\mathrm{d}x$$

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. -56 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

$$\frac{16x^{3}}{3}-28x^{2}+\int 49\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 49 integralini toping.

$$\frac{16x^{3}}{3}-28x^{2}+49x$$

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.

$$\frac{16x^{3}}{3}-28x^{2}+49x+С$$

x ga nisbatan hosilani topish

$$\left(4x-7\right)^{2}$$