Whakaoti i te ∫ (from 0 to 4) of x^3+x^2+x

$$\int _ { 0 } ^ { 4 } x ^ { 3 } + x ^ { 2 } + x$$

Aromātai

$$\frac{280}{3}\approx 93.333333333$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

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}.

$$\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x+\int x\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}+\int x\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}$$

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.

$$\frac{4^{4}}{4}+\frac{4^{3}}{3}+\frac{4^{2}}{2}-\left(\frac{0^{4}}{4}+\frac{0^{3}}{3}+\frac{0^{2}}{2}\right)$$

Whakarūnātia.

$$\frac{280}{3}$$