Whakaoti i te ∫ (from 0 to 1) of (x^2-3x+1) wrt x

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

Aromātai

$$-\frac{1}{6}\approx -0.166666667$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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}.

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

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 -3 ki te \frac{x^{2}}{2}.

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

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.

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

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.

$$\frac{1^{3}}{3}-\frac{3}{2}\times 1^{2}+1-\left(\frac{0^{3}}{3}-\frac{3}{2}\times 0^{2}+0\right)$$

Whakarūnātia.

$$-\frac{1}{6}$$