Whakaoti i te ∫ (from 0 to 1) of (6t^2+4t+6)(120t+40)dt

$$\int _ { 0 } ^ { 1 } ( 6 t ^ { 2 } + 4 t + 6 ) ( 120 t + 40 ) d t$$

Aromātai

$$1100$$

Ngā Upane Otinga

Whakamahia te āhuatanga tuaritanga hei whakarea te 6t^{2}+4t+6 ki te 120t+40 ka whakakotahi i ngā kupu rite.

$$\int _{0}^{1}720t^{3}+720t^{2}+880t+240\mathrm{d}t$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 720t^{3}+720t^{2}+880t+240\mathrm{d}t$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 720t^{3}\mathrm{d}t+\int 720t^{2}\mathrm{d}t+\int 880t\mathrm{d}t+\int 240\mathrm{d}t$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$720\int t^{3}\mathrm{d}t+720\int t^{2}\mathrm{d}t+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

Nā te mea \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int t^{3}\mathrm{d}t ki te \frac{t^{4}}{4}. Whakareatia 720 ki te \frac{t^{4}}{4}.

$$180t^{4}+720\int t^{2}\mathrm{d}t+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

Nā te mea \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int t^{2}\mathrm{d}t ki te \frac{t^{3}}{3}. Whakareatia 720 ki te \frac{t^{3}}{3}.

$$180t^{4}+240t^{3}+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

Nā te mea \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int t\mathrm{d}t ki te \frac{t^{2}}{2}. Whakareatia 880 ki te \frac{t^{2}}{2}.

$$180t^{4}+240t^{3}+440t^{2}+\int 240\mathrm{d}t$$

Kimihia te tau tōpū o 240 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}t=at.

$$180t^{4}+240t^{3}+440t^{2}+240t$$

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.

$$180\times 1^{4}+240\times 1^{3}+240\times 1+440\times 1^{2}-\left(180\times 0^{4}+240\times 0^{3}+240\times 0+440\times 0^{2}\right)$$

Whakarūnātia.

$$1100$$