Whakaoti i te ∫ (from 0 to sqrt { 2) of / 2 } (1/2x^2-x^4) wrt x

$$\int _ { 0 } ^ { \frac { \sqrt { 2 } } { 2 } } ( \frac { 1 } { 2 } x ^ { 2 } - x ^ { 4 } ) d x$$

Aromātai

$$\frac{\sqrt{2}}{60}\approx 0.023570226$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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

$$\frac{x^{3}}{6}-\int x^{4}\mathrm{d}x$$

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

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

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.

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

Whakarūnātia.

$$\frac{\sqrt{2}}{60}$$