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

$$\int _ { 1 } ^ { 2 } 4 x ^ { 2 } + x$$

Aromātai

$$\frac{65}{6}\approx 10.833333333$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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

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

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}.

$$\frac{4x^{3}}{3}+\frac{x^{2}}{2}$$

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.

$$\frac{4}{3}\times 2^{3}+\frac{2^{2}}{2}-\left(\frac{4}{3}\times 1^{3}+\frac{1^{2}}{2}\right)$$

Whakarūnātia.

$$\frac{65}{6}$$