Whakaoti i te ∫ sqrt{6}tdt

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

Aromātai

$$\frac{\sqrt{6}t^{2}}{2}+С$$

Ngā Upane Otinga

Whakatauwehetia te pūmau mā te whakamahi i te \int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t.

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

Nā te mea \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int t\mathrm{d}t ki te \frac{t^{2}}{2}.

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

Whakarūnātia.

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

Mēnā ko F\left(t\right) he pārōnaki kōaro o f\left(t\right), kāti ko te huinga o ngā pārōnaki kōaro katoa o f\left(t\right) ka whakaaturia e F\left(t\right)+C. Nō reira, me tāpiri te pūmau o te whakatōpūtanga C\in \mathrm{R} ki te otinga.

$$\frac{\sqrt{6}t^{2}}{2}+С$$

Kimi Pārōnaki e ai ki t

$$\sqrt{6}t$$