Whakaoti i te ∫ (2x^2-6x+1) wrt x

$$\int ( 2 x ^ { 2 } - 6 x + 1 ) d x$$

Aromātai

$$\frac{2x^{3}}{3}-3x^{2}+x+С$$

Ngā Upane Otinga

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 2x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 1\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$2\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{2}\mathrm{d}x ki te \frac{x^{3}}{3}. Whakareatia 2 ki te \frac{x^{3}}{3}.

$$\frac{2x^{3}}{3}-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x\mathrm{d}x ki te \frac{x^{2}}{2}. Whakareatia -6 ki te \frac{x^{2}}{2}.

$$\frac{2x^{3}}{3}-3x^{2}+\int 1\mathrm{d}x$$

Kimihia te tau tōpū o 1 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}-3x^{2}+x$$

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.

$$\frac{2x^{3}}{3}-3x^{2}+x+С$$

Kimi Pārōnaki e ai ki x

$$2x^{2}-6x+1$$