Whakaoti i te ∫ (4x-7)^2 wrt x

$$\int ( 4 x - 7 ) ^ { 2 } d x$$

Aromātai

$$\frac{\left(4x-7\right)^{3}}{12}+С$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(4x-7\right)^{2}.

$$\int 16x^{2}-56x+49\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 16x^{2}\mathrm{d}x+\int -56x\mathrm{d}x+\int 49\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$16\int x^{2}\mathrm{d}x-56\int x\mathrm{d}x+\int 49\mathrm{d}x$$

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 16 ki te \frac{x^{3}}{3}.

$$\frac{16x^{3}}{3}-56\int x\mathrm{d}x+\int 49\mathrm{d}x$$

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 -56 ki te \frac{x^{2}}{2}.

$$\frac{16x^{3}}{3}-28x^{2}+\int 49\mathrm{d}x$$

Kimihia te tau tōpū o 49 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{16x^{3}}{3}-28x^{2}+49x$$

Mēnā ko F\left(x\right) he pārōnaki kōaro o f\left(x\right), kāti ko te huinga o ngā pārōnaki kōaro katoa o f\left(x\right) ka whakaaturia e F\left(x\right)+C. Nō reira, me tāpiri te pūmau o te whakatōpūtanga C\in \mathrm{R} ki te otinga.

$$\frac{16x^{3}}{3}-28x^{2}+49x+С$$

Kimi Pārōnaki e ai ki x

$$\left(4x-7\right)^{2}$$