Whakaoti i te ∫ 6sqrt{x}

$$\int 6 \sqrt { x }$$

Aromātai

$$4x^{\frac{3}{2}}+С$$

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.

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

Tuhia anō te \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{3}{2}}}{\frac{3}{2}}. Whakarūnātia. Whakareatia 6 ki te \frac{2x^{\frac{3}{2}}}{3}.

$$4x^{\frac{3}{2}}$$

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.

$$4x^{\frac{3}{2}}+С$$

Kimi Pārōnaki e ai ki x

$$6\sqrt{x}$$