Whakaoti i te ∫ (from 1 to 4) of left(-x+4/xright) wrt x

$$\int_{ 1 }^{ 4 } \left( -x+ \frac{ 4 }{ x } \right) d x$$

Aromātai

$$8\ln(2)-\frac{15}{2}\approx -1.954822556$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$-\int x\mathrm{d}x+4\int \frac{1}{x}\mathrm{d}x$$

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

$$-\frac{x^{2}}{2}+4\int \frac{1}{x}\mathrm{d}x$$

Whakamahia te \int \frac{1}{x}\mathrm{d}x=\ln(|x|) mai i te ripanga o ngā tau tōpū pātahi kia whakaputa i te huanga.

$$-\frac{x^{2}}{2}+4\ln(|x|)$$

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.

$$-\frac{4^{2}}{2}+4\ln(|4|)-\left(-\frac{1^{2}}{2}+4\ln(|1|)\right)$$

Whakarūnātia.

$$-\frac{15}{2}+8\ln(2)$$