Réitigh ∫ (4-5x)^3 wrt x

$$\int ( 4 - 5 x ) ^ { 3 } d x$$

Luacháil

$$-\frac{\left(4-5x\right)^{4}}{20}+С$$

Céimeanna Réitigh

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} chun \left(4-5x\right)^{3} a leathnú.

$$\int 64-240x+300x^{2}-125x^{3}\mathrm{d}x$$

Measc an tsuim téarma fá téarma.

$$\int 64\mathrm{d}x+\int -240x\mathrm{d}x+\int 300x^{2}\mathrm{d}x+\int -125x^{3}\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$\int 64\mathrm{d}x-240\int x\mathrm{d}x+300\int x^{2}\mathrm{d}x-125\int x^{3}\mathrm{d}x$$

Aimsigh suimeálaithe do 64 ag baint úsáid as an tábla do suimeálaithe coitianta riail\int a\mathrm{d}x=ax.

$$64x-240\int x\mathrm{d}x+300\int x^{2}\mathrm{d}x-125\int x^{3}\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x\mathrm{d}x le \frac{x^{2}}{2}. Méadaigh -240 faoi \frac{x^{2}}{2}.

$$64x-120x^{2}+300\int x^{2}\mathrm{d}x-125\int x^{3}\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{2}\mathrm{d}x le \frac{x^{3}}{3}. Méadaigh 300 faoi \frac{x^{3}}{3}.

$$64x-120x^{2}+100x^{3}-125\int x^{3}\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{3}\mathrm{d}x le \frac{x^{4}}{4}. Méadaigh -125 faoi \frac{x^{4}}{4}.

$$64x-120x^{2}+100x^{3}-\frac{125x^{4}}{4}$$

Má tá F\left(x\right) mar frithdhíorthach do f\left(x\right), beidh tacar do frithdhíorthach uile do f\left(x\right) a thabhairt ag F\left(x\right)+C. Mar sin de, cur an comhtháthú leanúnach C\in \mathrm{R} don toradh.

$$64x-120x^{2}+100x^{3}-\frac{125x^{4}}{4}+С$$

Difreálaigh w.r.t. x

$$\left(4-5x\right)^{3}$$