Réitigh ∫ (from 0 to 2) of (4x-x^3)^2 wrt x

$$\int _ { 0 } ^ { 2 } ( 4 x - x ^ { 3 } ) ^ { 2 } d x$$

Luacháil

$$\frac{1024}{105}\approx 9.752380952$$

Céimeanna Réitigh

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

$$\int _{0}^{2}16x^{2}-8xx^{3}+\left(x^{3}\right)^{2}\mathrm{d}x$$

Chun cumhachtaí den bhonn céanna a iolrú, suimigh a n-easpónaint. Suimigh 1 agus 3 chun 4 a bhaint amach.

$$\int _{0}^{2}16x^{2}-8x^{4}+\left(x^{3}\right)^{2}\mathrm{d}x$$

Chun cumhacht a ardú go cumhacht eile, méadaigh na heaspónaint. Iolraigh 3 agus 2 chun 6 a bhaint amach.

$$\int _{0}^{2}16x^{2}-8x^{4}+x^{6}\mathrm{d}x$$

Déan luacháil ar an suimeálaí éiginnte ar dtús.

$$\int 16x^{2}-8x^{4}+x^{6}\mathrm{d}x$$

Measc an tsuim téarma fá téarma.

$$\int 16x^{2}\mathrm{d}x+\int -8x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$16\int x^{2}\mathrm{d}x-8\int x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

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 16 faoi \frac{x^{3}}{3}.

$$\frac{16x^{3}}{3}-8\int x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{4}\mathrm{d}x le \frac{x^{5}}{5}. Méadaigh -8 faoi \frac{x^{5}}{5}.

$$\frac{16x^{3}}{3}-\frac{8x^{5}}{5}+\int x^{6}\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{6}\mathrm{d}x le \frac{x^{7}}{7}.

$$\frac{16x^{3}}{3}-\frac{8x^{5}}{5}+\frac{x^{7}}{7}$$

Simpligh.

$$\frac{x^{7}}{7}-\frac{8x^{5}}{5}+\frac{16x^{3}}{3}$$

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

$$\frac{2^{7}}{7}-\frac{8}{5}\times 2^{5}+\frac{16}{3}\times 2^{3}-\left(\frac{0^{7}}{7}-\frac{8}{5}\times 0^{5}+\frac{16}{3}\times 0^{3}\right)$$

Simpligh.

$$\frac{1024}{105}$$