Whakaoti i te ∫ _{frac{sqrt{2}}{2}}^1(x^4-1/2x^2) wrt x

$$\int _ { \frac { \sqrt { 2 } } { 2 } } ^ { 1 } ( x ^ { 4 } - \frac { 1 } { 2 } x ^ { 2 } ) d x$$

Aromātai

$$\frac{\sqrt{2}}{60}+\frac{1}{30}\approx 0.056903559$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int x^{4}-\frac{x^{2}}{2}\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$\int x^{4}\mathrm{d}x-\frac{\int x^{2}\mathrm{d}x}{2}$$

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{4}\mathrm{d}x ki te \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-\frac{\int x^{2}\mathrm{d}x}{2}$$

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 -\frac{1}{2} ki te \frac{x^{3}}{3}.

$$\frac{x^{5}}{5}-\frac{x^{3}}{6}$$

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.

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

Whakarūnātia.

$$\frac{1}{30}+\frac{\sqrt{2}}{60}$$