Réitigh ∫ 1/4x^3-1/3x^2+1/2x wrt x

$$\int{ \frac{ 1 }{ 4 } { x }^{ 3 } - \frac{ 1 }{ 3 } { x }^{ 2 } + \frac{ 1 }{ 2 } x }d x$$

Luacháil

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}+С$$

Céimeanna Réitigh

Measc an tsuim téarma fá téarma.

$$\int \frac{x^{3}}{4}\mathrm{d}x+\int -\frac{x^{2}}{3}\mathrm{d}x+\int \frac{x}{2}\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$\frac{\int x^{3}\mathrm{d}x}{4}-\frac{\int x^{2}\mathrm{d}x}{3}+\frac{\int x\mathrm{d}x}{2}$$

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 \frac{1}{4} faoi \frac{x^{4}}{4}.

$$\frac{x^{4}}{16}-\frac{\int x^{2}\mathrm{d}x}{3}+\frac{\int x\mathrm{d}x}{2}$$

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 -\frac{1}{3} faoi \frac{x^{3}}{3}.

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{\int x\mathrm{d}x}{2}$$

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 \frac{1}{2} faoi \frac{x^{2}}{2}.

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}$$

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.

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}+С$$

Difreálaigh w.r.t. x

$$\frac{x^{3}}{4}-\frac{x^{2}}{3}+\frac{x}{2}$$