Whakaoti i te ∫ (from 5 to 10) of x+sin(x)+12 wrt x

$$\int_{ 5 }^{ 10 } x+ \sin ( x ) +12 d x$$

Aromātai

$$\frac{2\cos(5)+195-2\cos(10)}{2}\approx 98.622733715$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int x+\sin(x)+12\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x\mathrm{d}x+\int \sin(x)\mathrm{d}x+\int 12\mathrm{d}x$$

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}.

$$\frac{x^{2}}{2}+\int \sin(x)\mathrm{d}x+\int 12\mathrm{d}x$$

Whakamahia te \int \sin(x)\mathrm{d}x=-\cos(x) mai i te ripanga o ngā tau tōpū pātahi kia whakaputa i te huanga.

$$\frac{x^{2}}{2}-\cos(x)+\int 12\mathrm{d}x$$

Kimihia te tau tōpū o 12 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{2}-\cos(x)+12x$$

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.

$$\frac{10^{2}}{2}-\cos(10)+10\times 12-\left(\frac{5^{2}}{2}-\cos(5)+5\times 12\right)$$

Whakarūnātia.

$$\frac{1}{2}\left(-2\cos(10)+195+2\cos(5)\right)$$