Whakaoti i te ∫ (from 0 to 1) of (1-8v^3+16v^7)dv

$$\int _ { 0 } ^ { 1 } ( 1 - 8 v ^ { 3 } + 16 v ^ { 7 } ) d v$$

Aromātai

$$1$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 1-8v^{3}+16v^{7}\mathrm{d}v$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 1\mathrm{d}v+\int -8v^{3}\mathrm{d}v+\int 16v^{7}\mathrm{d}v$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$\int 1\mathrm{d}v-8\int v^{3}\mathrm{d}v+16\int v^{7}\mathrm{d}v$$

Kimihia te tau tōpū o 1 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}v=av.

$$v-8\int v^{3}\mathrm{d}v+16\int v^{7}\mathrm{d}v$$

Nā te mea \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int v^{3}\mathrm{d}v ki te \frac{v^{4}}{4}. Whakareatia -8 ki te \frac{v^{4}}{4}.

$$v-2v^{4}+16\int v^{7}\mathrm{d}v$$

Nā te mea \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int v^{7}\mathrm{d}v ki te \frac{v^{8}}{8}. Whakareatia 16 ki te \frac{v^{8}}{8}.

$$v-2v^{4}+2v^{8}$$

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.

$$1-2\times 1^{4}+2\times 1^{8}-\left(0-2\times 0^{4}+2\times 0^{8}\right)$$

Whakarūnātia.

$$1$$