Réitigh ∫ sqrt{6}tdt

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

Luacháil

$$\frac{\sqrt{6}t^{2}}{2}+С$$

Céimeanna Réitigh

Fág an úsáid leanúnach do\int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t as an áireamh.

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

Ó \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int t\mathrm{d}t le \frac{t^{2}}{2}.

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

Simpligh.

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

Má tá F\left(t\right) mar frithdhíorthach do f\left(t\right), beidh tacar do frithdhíorthach uile do f\left(t\right) a thabhairt ag F\left(t\right)+C. Mar sin de, cur an comhtháthú leanúnach C\in \mathrm{R} don toradh.

$$\frac{\sqrt{6}t^{2}}{2}+С$$

Difreálaigh w.r.t. t

$$\sqrt{6}t$$