∫ sqrt{6}tdt eching

$$\int \sqrt { 6 } t d t$$

Baholash

$$\frac{\sqrt{6}t^{2}}{2}+С$$

Yechim qadamlari

\int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t yordamidan konstantani qavsdan tashqariga chiqaring.

$$\sqrt{6}\int t\mathrm{d}t$$

k\neq -1 uchun integral \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} boʻlgani uchun, \int t\mathrm{d}t integralni \frac{t^{2}}{2} bilan almashtiring.

$$\sqrt{6}\times \frac{t^{2}}{2}$$

Qisqartirish.

$$\frac{\sqrt{6}t^{2}}{2}$$

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

$$\frac{\sqrt{6}t^{2}}{2}+С$$

t ga nisbatan hosilani topish

$$\sqrt{6}t$$