Whakaoti i te ∫ (from 0 to 1) of (2x+3)*(3x-5) wrt x

$$\int _ { 0 } ^ { 1 } ( 2 x + 3 ) \cdot ( 3 x - 5 ) d x$$

Aromātai

$$-\frac{27}{2}=-13.5$$

Ngā Upane Otinga

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 2x+3 ki ia tau o 3x-5.

$$\int _{0}^{1}6x^{2}-10x+9x-15\mathrm{d}x$$

Pahekotia te -10x me 9x, ka -x.

$$\int _{0}^{1}6x^{2}-x-15\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 6x^{2}-x-15\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 6x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int -15\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$6\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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}. Whakareatia 6 ki te \frac{x^{3}}{3}.

$$2x^{3}-\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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 -1 ki te \frac{x^{2}}{2}.

$$2x^{3}-\frac{x^{2}}{2}+\int -15\mathrm{d}x$$

Kimihia te tau tōpū o -15 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$2x^{3}-\frac{x^{2}}{2}-15x$$

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.

$$2\times 1^{3}-\frac{1^{2}}{2}-15-\left(2\times 0^{3}-\frac{0^{2}}{2}-15\times 0\right)$$

Whakarūnātia.

$$-\frac{27}{2}$$