Whakaoti i te ∫ (from 6 to 10) of (-x^3/3+14733x)(06x) wrt x

$$\int_{ 6 }^{ 10 } (- \frac{ { x }^{ 3 } }{ 3 } +14733x)(06x) d x$$

Aromātai

$$0$$

Ngā Upane Otinga

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 14733x ki te \frac{3}{3}.

$$\int _{6}^{10}\left(-\frac{x^{3}}{3}+\frac{3\times 14733x}{3}\right)\times 0\times 6x\mathrm{d}x$$

Tā te mea he rite te tauraro o -\frac{x^{3}}{3} me \frac{3\times 14733x}{3}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\int _{6}^{10}\frac{-x^{3}+3\times 14733x}{3}\times 0\times 6x\mathrm{d}x$$

Mahia ngā whakarea i roto o -x^{3}+3\times 14733x.

$$\int _{6}^{10}\frac{-x^{3}+44199x}{3}\times 0\times 6x\mathrm{d}x$$

Whakareatia te 0 ki te 6, ka 0.

$$\int _{6}^{10}\frac{-x^{3}+44199x}{3}\times 0x\mathrm{d}x$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$\int _{6}^{10}0\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 0\mathrm{d}x$$

Kimihia te tau tōpū o 0 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$0$$

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.

$$0+0$$

Whakarūnātia.

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