Whakaoti i te ∫ (from 10 to 20) of (x^2-1)e^-02x wrt x

$$\int _ { 10 } ^ { 20 } ( x ^ { 2 } - 1 ) e ^ { - 02 x } d x$$

Aromātai

$$\frac{6970}{3}\approx 2323.333333333$$

Ngā Upane Otinga

Whakareatia te 0 ki te 2, ka 0.

$$\int _{10}^{20}\left(x^{2}-1\right)e^{0x}\mathrm{d}x$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$\int _{10}^{20}\left(x^{2}-1\right)e^{0}\mathrm{d}x$$

Tātaihia te e mā te pū o 0, kia riro ko 1.

$$\int _{10}^{20}\left(x^{2}-1\right)\times 1\mathrm{d}x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-1 ki te 1.

$$\int _{10}^{20}x^{2}-1\mathrm{d}x$$

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

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}.

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

Kimihia te tau tōpū o -1 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

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

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.

$$\frac{20^{3}}{3}-20-\left(\frac{10^{3}}{3}-10\right)$$

Whakarūnātia.

$$\frac{6970}{3}$$