Whakaoti i te ∫ (from 0 to 1) of (x^2+sinx) wrt x

$$\int _ { 0 } ^ { 1 } ( x ^ { 2 } + \sin x ) d x$$

Aromātai

$$\frac{4-3\cos(1)}{3}\approx 0.793031027$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{2}\mathrm{d}x ki te \frac{x^{3}}{3}.

$$\frac{x^{3}}{3}+\int \sin(x)\mathrm{d}x$$

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

$$\frac{x^{3}}{3}-\cos(x)$$

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.

$$\frac{1^{3}}{3}-\cos(1)-\left(\frac{0^{3}}{3}-\cos(0)\right)$$

Whakarūnātia.

$$\frac{1}{3}\left(4-3\cos(1)\right)$$