Whakaoti i te ∫ (from 2 to 10) of (x^3+4x+9)

$$\int _ { 2 } ^ { 10 } ( x ^ { 3 } + 4 x + 9 )$$

Aromātai

$$2760$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int x^{3}+4x+9\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x^{3}\mathrm{d}x+\int 4x\mathrm{d}x+\int 9\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$\int x^{3}\mathrm{d}x+4\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+4\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+2x^{2}+\int 9\mathrm{d}x$$

Kimihia te tau tōpū o 9 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}+2x^{2}+9x$$

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.

$$\frac{10^{4}}{4}+2\times 10^{2}+9\times 10-\left(\frac{2^{4}}{4}+2\times 2^{2}+9\times 2\right)$$

Whakarūnātia.

$$2760$$