Whakaoti i te ∫ (from 0 to log ( 1+ sqrt{ 2) of )}{left(e^x-e^x/2right)}^3{left(e^x-e^x/2right)}^11 wrt x

$$\int_{ 0 }^{ \log ( 1+ \sqrt{ 2 } ) } { \left( \frac{ { e }^{ x } - { e }^{ x } }{ 2 } \right) }^{ 3 } { \left( \frac{ { e }^{ x } - { e }^{ x } }{ 2 } \right) }^{ 11 } d x$$

Aromātai

$$0$$

Ngā Upane Otinga

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 3 me te 11 kia riro ai te 14.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}\left(\frac{e^{x}-e^{x}}{2}\right)^{14}\mathrm{d}x$$

Pahekotia te e^{x} me -e^{x}, ka 0.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}\left(\frac{0}{2}\right)^{14}\mathrm{d}x$$

Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}0^{14}\mathrm{d}x$$

Tātaihia te 0 mā te pū o 14, kia riro ko 0.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}0\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

$$\int 0\mathrm{d}x$$

Kimihia te tau tōpū o 0 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

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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.

$$0+0$$

Whakarūnātia.

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