Whakaoti i te ∫ (from 0 to 1) of 5u^5+3u^2+udu

$$\int _ { 0 } ^ { 1 } 5 u ^ { 5 } + 3 u ^ { 2 } + u d u$$

Aromātai

$$\frac{7}{3}\approx 2.333333333$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 5u^{5}+3u^{2}+u\mathrm{d}u$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 5u^{5}\mathrm{d}u+\int 3u^{2}\mathrm{d}u+\int u\mathrm{d}u$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$5\int u^{5}\mathrm{d}u+3\int u^{2}\mathrm{d}u+\int u\mathrm{d}u$$

Nā te mea \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int u^{5}\mathrm{d}u ki te \frac{u^{6}}{6}. Whakareatia 5 ki te \frac{u^{6}}{6}.

$$\frac{5u^{6}}{6}+3\int u^{2}\mathrm{d}u+\int u\mathrm{d}u$$

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

$$\frac{5u^{6}}{6}+u^{3}+\int u\mathrm{d}u$$

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

$$\frac{5u^{6}}{6}+u^{3}+\frac{u^{2}}{2}$$

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.

$$\frac{5}{6}\times 1^{6}+1^{3}+\frac{1^{2}}{2}-\left(\frac{5}{6}\times 0^{6}+0^{3}+\frac{0^{2}}{2}\right)$$

Whakarūnātia.

$$\frac{7}{3}$$