Whakaoti i te ∫ 1/2sqrt{x}

$$\int \frac { 1 } { 2 \sqrt { x } }$$

Aromātai

$$\sqrt{x}+С$$

Ngā Upane Otinga

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

$$\frac{\int \frac{1}{\sqrt{x}}\mathrm{d}x}{2}$$

Tuhia anō te \frac{1}{\sqrt{x}} hei x^{-\frac{1}{2}}. Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{-\frac{1}{2}}\mathrm{d}x ki te \frac{x^{\frac{1}{2}}}{\frac{1}{2}}. Whakarūnāhia me te tahuri mai i te āhua taupū ki te āhua pūtake.

$$\sqrt{x}$$

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

$$\sqrt{x}+С$$

Kimi Pārōnaki e ai ki x

$$\frac{1}{2\sqrt{x}}$$