Whakaoti i te ∫ (from 4 to 9) of (sqrt{x}+1)sqrt{x wrt x

$$\int _ { 4 } ^ { 9 } ( \sqrt { x } + 1 ) \sqrt { x } d x$$

Aromātai

$$\frac{271}{6}\approx 45.166666667$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te \sqrt{x}+1 ki te \sqrt{x}.

$$\int _{4}^{9}\left(\sqrt{x}\right)^{2}+\sqrt{x}\mathrm{d}x$$

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

$$\int _{4}^{9}x+\sqrt{x}\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int x+\sqrt{x}\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x\mathrm{d}x+\int \sqrt{x}\mathrm{d}x$$

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}.

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

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.

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

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.

$$\frac{9^{2}}{2}+\frac{2}{3}\times 9^{\frac{3}{2}}-\left(\frac{4^{2}}{2}+\frac{2}{3}\times 4^{\frac{3}{2}}\right)$$

Whakarūnātia.

$$\frac{271}{6}$$