Whakaoti i te ∫ (from 0.2 to 12) of (-800V+1160)dV

$$\int _ { 0.2 } ^ { 12 } ( - 800 V + 1160 ) d V$$

Aromātai

$$-43896$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int -800V+1160\mathrm{d}V$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int -800V\mathrm{d}V+\int 1160\mathrm{d}V$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$-800\int V\mathrm{d}V+\int 1160\mathrm{d}V$$

Nā te mea \int V^{k}\mathrm{d}V=\frac{V^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int V\mathrm{d}V ki te \frac{V^{2}}{2}. Whakareatia -800 ki te \frac{V^{2}}{2}.

$$-400V^{2}+\int 1160\mathrm{d}V$$

Kimihia te tau tōpū o 1160 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}V=aV.

$$-400V^{2}+1160V$$

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.

$$-400\times 12^{2}+1160\times 12-\left(-400\times 0.2^{2}+1160\times 0.2\right)$$

Whakarūnātia.

$$-43896$$