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

$$\int_{ 0 }^{ 5 } (x+2- \frac{ x }{ 2 } ) d x$$

Aromātai

$$\frac{65}{4}=16.25$$

Ngā Upane Otinga

Pahekotia te x me -\frac{x}{2}, ka \frac{1}{2}x.

$$\int _{0}^{5}\frac{1}{2}x+2\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x\mathrm{d}x ki te \frac{x^{2}}{2}. Whakareatia \frac{1}{2} ki te \frac{x^{2}}{2}.

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

Kimihia te tau tōpū o 2 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{4}+2x$$

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.

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

Whakarūnātia.

$$\frac{65}{4}$$