Whakaoti i te ∫ (from 0 to 1) of x^2left(x-4right)^2 wrt x

$$\int_{ 0 }^{ 1 } { x }^{ 2 } { \left(x-4 \right) }^{ 2 } d x$$

Aromātai

$$\frac{53}{15}\approx 3.533333333$$

Ngā Upane Otinga

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

$$\int _{0}^{1}x^{2}\left(x^{2}-8x+16\right)\mathrm{d}x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2} ki te x^{2}-8x+16.

$$\int _{0}^{1}x^{4}-8x^{3}+16x^{2}\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int x^{4}-8x^{3}+16x^{2}\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x^{4}\mathrm{d}x+\int -8x^{3}\mathrm{d}x+\int 16x^{2}\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$\int x^{4}\mathrm{d}x-8\int x^{3}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{4}\mathrm{d}x ki te \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-8\int x^{3}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

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}. Whakareatia -8 ki te \frac{x^{4}}{4}.

$$\frac{x^{5}}{5}-2x^{4}+16\int x^{2}\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-2x^{4}+\frac{16x^{3}}{3}$$

Whakarūnātia.

$$\frac{16x^{3}}{3}-2x^{4}+\frac{x^{5}}{5}$$

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.

$$\frac{16}{3}\times 1^{3}-2\times 1^{4}+\frac{1^{5}}{5}-\left(\frac{16}{3}\times 0^{3}-2\times 0^{4}+\frac{0^{5}}{5}\right)$$

Whakarūnātia.

$$\frac{53}{15}$$