Whakaoti i te ∫ (from - 2 to 5) of (4x-3)^2 wrt x

$$\int _ { - 2 } ^ { 5 } ( 4 x - 3 ) ^ { 2 } d x$$

Aromātai

$$\frac{1561}{3}\approx 520.333333333$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(4x-3\right)^{2}.

$$\int _{-2}^{5}16x^{2}-24x+9\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 16x^{2}-24x+9\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 16x^{2}\mathrm{d}x+\int -24x\mathrm{d}x+\int 9\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$16\int x^{2}\mathrm{d}x-24\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 16 ki te \frac{x^{3}}{3}.

$$\frac{16x^{3}}{3}-24\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 -24 ki te \frac{x^{2}}{2}.

$$\frac{16x^{3}}{3}-12x^{2}+\int 9\mathrm{d}x$$

Kimihia te tau tōpū o 9 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{16x^{3}}{3}-12x^{2}+9x$$

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.

$$\frac{16}{3}\times 5^{3}-12\times 5^{2}+9\times 5-\left(\frac{16}{3}\left(-2\right)^{3}-12\left(-2\right)^{2}+9\left(-2\right)\right)$$

Whakarūnātia.

$$\frac{1561}{3}$$