Whakaoti i te ∫ (from -1 to 3) of x(x-1)(x+2) wrt x

$$\int_{ -1 }^{ 3 } x(x-1)(x+2) d x$$

Aromātai

$$\frac{64}{3}\approx 21.333333333$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x-1.

$$\int _{-1}^{3}\left(x^{2}-x\right)\left(x+2\right)\mathrm{d}x$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o x^{2}-x ki ia tau o x+2.

$$\int _{-1}^{3}x^{3}+2x^{2}-x^{2}-2x\mathrm{d}x$$

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

$$\int _{-1}^{3}x^{3}+x^{2}-2x\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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}.

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

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}.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}-2\int x\mathrm{d}x$$

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}. Whakareatia -2 ki te \frac{x^{2}}{2}.

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

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.

$$\frac{3^{4}}{4}+\frac{3^{3}}{3}-3^{2}-\left(\frac{\left(-1\right)^{4}}{4}+\frac{\left(-1\right)^{3}}{3}-\left(-1\right)^{2}\right)$$

Whakarūnātia.

$$\frac{64}{3}$$