Whakaoti i te ∫ (from 1 to 4) of (4x^2+3) wrt x

$$\int _ { 1 } ^ { 4 } ( 4 x ^ { 2 } + 3 ) d x$$

Aromātai

$$93$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 4x^{2}+3\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 4x^{2}\mathrm{d}x+\int 3\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$4\int x^{2}\mathrm{d}x+\int 3\mathrm{d}x$$

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}. Whakareatia 4 ki te \frac{x^{3}}{3}.

$$\frac{4x^{3}}{3}+\int 3\mathrm{d}x$$

Kimihia te tau tōpū o 3 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{4x^{3}}{3}+3x$$

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.

$$\frac{4}{3}\times 4^{3}+3\times 4-\left(\frac{4}{3}\times 1^{3}+3\times 1\right)$$

Whakarūnātia.

$$93$$