Whakaoti i te f ( x ) = ∫ (from 0 to x) of (t^3+2t^2+1)dt

$$f ( x ) = \int _ { 0 } ^ { x } ( t ^ { 3 } + 2 t ^ { 2 } + 1 ) d t$$

Aromātai

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

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

Nā te mea \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int t^{3}\mathrm{d}t ki te \frac{t^{4}}{4}.

$$\frac{t^{4}}{4}+2\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

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

$$\frac{t^{4}}{4}+\frac{2t^{3}}{3}+\int 1\mathrm{d}t$$

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

$$\frac{t^{4}}{4}+\frac{2t^{3}}{3}+t$$

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.

$$\frac{x^{4}}{4}+\frac{2}{3}x^{3}+x-\left(\frac{0^{4}}{4}+\frac{2}{3}\times 0^{3}+0\right)$$

Whakarūnātia.

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

Kimi Pārōnaki e ai ki x

$$x^{3}+2x^{2}+1$$