Whakaoti i te F ( x ) = ∫ (from 1 to x) of (t^2-4)dt

$$F ( x ) = \int _ { 1 } ^ { x } ( t ^ { 2 } - 4 ) d t$$

Aromātai

$$\frac{x^{3}-12x+11}{3}$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int t^{2}-4\mathrm{d}t$$

Kōmitimititia te kīanga tapeke mā te kīanga.

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

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}.

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

Kimihia te tau tōpū o -4 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}t=at.

$$\frac{t^{3}}{3}-4t$$

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.

$$\frac{x^{3}}{3}-4x-\left(\frac{1^{3}}{3}-4\right)$$

Whakarūnātia.

$$\frac{x^{3}}{3}-4x+\frac{11}{3}$$

Kimi Pārōnaki e ai ki x

$$x^{2}-4$$