Réitigh ∫ (from 0 to 4) of (4.4x-4)(-0.2-0.1x) wrt x

$$\int_{ 0 }^{ 4 } (4.4x-4)(-0.2-0.1x) d x$$

Luacháil

$$-\frac{752}{75}\approx -10.026666667$$

Céimeanna Réitigh

Cuir an t-airí dáileacháin i bhfeidhm trí gach téarma de 4.4x-4 a iolrú faoi gach téarma de -0.2-0.1x.

$$\int _{0}^{4}-0.88x-0.44x^{2}+0.8+0.4x\mathrm{d}x$$

Comhcheangail -0.88x agus 0.4x chun -0.48x a fháil.

$$\int _{0}^{4}-0.48x-0.44x^{2}+0.8\mathrm{d}x$$

Déan luacháil ar an suimeálaí éiginnte ar dtús.

$$\int -\frac{12x}{25}-\frac{11x^{2}}{25}+0.8\mathrm{d}x$$

Measc an tsuim téarma fá téarma.

$$\int -\frac{12x}{25}\mathrm{d}x+\int -\frac{11x^{2}}{25}\mathrm{d}x+\int 0.8\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$-\frac{12\int x\mathrm{d}x}{25}-\frac{11\int x^{2}\mathrm{d}x}{25}+\int 0.8\mathrm{d}x$$

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 -0.48 faoi \frac{x^{2}}{2}.

$$-\frac{6x^{2}}{25}-\frac{11\int x^{2}\mathrm{d}x}{25}+\int 0.8\mathrm{d}x$$

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 -0.44 faoi \frac{x^{3}}{3}.

$$-\frac{6x^{2}}{25}-\frac{11x^{3}}{75}+\int 0.8\mathrm{d}x$$

Aimsigh suimeálaithe do 0.8 ag baint úsáid as an tábla do suimeálaithe coitianta riail\int a\mathrm{d}x=ax.

$$-\frac{6x^{2}}{25}-\frac{11x^{3}}{75}+\frac{4x}{5}$$

Is ionann suimeálaí cinnte agus frithdhíorthach an nath luacháilte ag teorainn uachtair na suimeála lúide an frithdhíorthach luacháilte ag teorainn íochtair na suimeála.

$$-\frac{6}{25}\times 4^{2}-\frac{11}{75}\times 4^{3}+0.8\times 4-\left(-\frac{6}{25}\times 0^{2}-\frac{11}{75}\times 0^{3}+0.8\times 0\right)$$

Simpligh.

$$-\frac{752}{75}$$