∫ (x+3)(x-4) wrt x eching

$$\int ( x + 3 ) ( x - 4 ) d x$$

Baholash

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-12x+С$$

Yechim qadamlari

x+3 ifodaning har bir elementini x-4 ifodaning har bir elementiga ko‘paytirish orqali taqsimot qonuni xususiyatlarini qo‘llash mumkin.

$$\int x^{2}-4x+3x-12\mathrm{d}x$$

-x ni olish uchun -4x va 3x ni birlashtirish.

$$\int x^{2}-x-12\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int -12\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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.

$$\frac{x^{3}}{3}-\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}+\int -12\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, -12 integralini toping.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-12x$$

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.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-12x+С$$

x ga nisbatan hosilani topish

$$\left(x-4\right)\left(x+3\right)$$