Whakaoti i te ∫ (from - 3 to 3) of (4x^2-2x+1) wrt x

$$\int _ { - 3 } ^ { 3 } ( 4 x ^ { 2 } - 2 x + 1 ) d x$$

Aromātai

$$78$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 4x^{2}-2x+1\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 4x^{2}\mathrm{d}x+\int -2x\mathrm{d}x+\int 1\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$4\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

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

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}.

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

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.

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

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.

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

Whakarūnātia.

$$78$$