Whakaoti i te ∫ (from 1 to 3) of (3x^2+2x+1) wrt x

$$\int _ { 1 } ^ { 3 } ( 3 x ^ { 2 } + 2 x + 1 ) d x$$

Aromātai

$$36$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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 3 ki te \frac{x^{3}}{3}.

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

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 2 ki te \frac{x^{2}}{2}.

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

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.

$$x^{3}+x^{2}+x$$

Ko te tau tōpū tautuhi ko te pārōnaki kōaro o te kīanga i aromātaitia i te tepe tōrunga o te pāwhaitua, tangohia te pārōnaki kōaro i aromātaitia i te tepe tōraro o te pāwhaitua.

$$3^{3}+3^{2}+3-\left(1^{3}+1^{2}+1\right)$$

Whakarūnātia.

$$36$$