Whakaoti i te ∫ (from 0 to 4) of (6-(4-sqrt{x})^2) wrt x

$$\int _ { 0 } ^ { 4 } ( 6 - ( 4 - \sqrt { x } ) ^ { 2 } ) d x$$

Aromātai

$$-\frac{16}{3}\approx -5.333333333$$

Ngā Upane Otinga

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

$$\int _{0}^{4}6-\left(16-8\sqrt{x}+\left(\sqrt{x}\right)^{2}\right)\mathrm{d}x$$

Tātaihia te \sqrt{x} mā te pū o 2, kia riro ko x.

$$\int _{0}^{4}6-\left(16-8\sqrt{x}+x\right)\mathrm{d}x$$

Hei kimi i te tauaro o 16-8\sqrt{x}+x, kimihia te tauaro o ia taurangi.

$$\int _{0}^{4}6-16+8\sqrt{x}-x\mathrm{d}x$$

Tangohia te 16 i te 6, ka -10.

$$\int _{0}^{4}-10+8\sqrt{x}-x\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int -10+8\sqrt{x}-x\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int -10\mathrm{d}x+\int 8\sqrt{x}\mathrm{d}x+\int -x\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$\int -10\mathrm{d}x+8\int \sqrt{x}\mathrm{d}x-\int x\mathrm{d}x$$

Kimihia te tau tōpū o -10 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$-10x+8\int \sqrt{x}\mathrm{d}x-\int x\mathrm{d}x$$

Tuhia anō te \sqrt{x} hei x^{\frac{1}{2}}. Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{\frac{1}{2}}\mathrm{d}x ki te \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Whakarūnātia. Whakareatia 8 ki te \frac{2x^{\frac{3}{2}}}{3}.

$$-10x+\frac{16x^{\frac{3}{2}}}{3}-\int x\mathrm{d}x$$

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

$$-10x+\frac{16x^{\frac{3}{2}}}{3}-\frac{x^{2}}{2}$$

Whakarūnātia.

$$-10x-\frac{x^{2}}{2}+\frac{16x^{\frac{3}{2}}}{3}$$

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.

$$-10\times 4-\frac{4^{2}}{2}+\frac{16}{3}\times 4^{\frac{3}{2}}-\left(-10\times 0-\frac{0^{2}}{2}+\frac{16}{3}\times 0^{\frac{3}{2}}\right)$$

Whakarūnātia.

$$-\frac{16}{3}$$