Whakaoti i te ∫ (from 0 to 17) of (9x^2+48x^3)

$$\int _ { 0 } ^ { 17 } ( 9 x ^ { 2 } + 48 x ^ { 3 } )$$

Aromātai

$$1016991$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 9x^{2}+48x^{3}\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 9x^{2}\mathrm{d}x+\int 48x^{3}\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$9\int x^{2}\mathrm{d}x+48\int x^{3}\mathrm{d}x$$

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

$$3x^{3}+48\int x^{3}\mathrm{d}x$$

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}. Whakareatia 48 ki te \frac{x^{4}}{4}.

$$3x^{3}+12x^{4}$$

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.

$$3\times 17^{3}+12\times 17^{4}-\left(3\times 0^{3}+12\times 0^{4}\right)$$

Whakarūnātia.

$$1016991$$