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

$$\int_{ 1 }^{ 4 } { x }^{ 2 } + \frac{ 1 }{ { x }^{ 3 } } d x$$

Aromātai

$$\frac{687}{32}=21.46875$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int x^{2}+\frac{1}{x^{3}}\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x^{2}\mathrm{d}x+\int \frac{1}{x^{3}}\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+\int \frac{1}{x^{3}}\mathrm{d}x$$

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

$$\frac{x^{3}}{3}-\frac{1}{2x^{2}}$$

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.

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

Whakarūnātia.

$$\frac{687}{32}$$