Réitigh ∫ (from 1 to 2) of (x+sqrt[3]{x}+1/x^2) wrt x

$$\int _ { 1 } ^ { 2 } ( x + \sqrt[ 3 ] { x } + \frac { 1 } { x ^ { 2 } } ) d x$$

Luacháil

$$\frac{3\sqrt[3]{2}}{2}+\frac{5}{4}\approx 3.139881575$$

Céimeanna Réitigh

Déan luacháil ar an suimeálaí éiginnte ar dtús.

$$\int x+\sqrt[3]{x}+\frac{1}{x^{2}}\mathrm{d}x$$

Measc an tsuim téarma fá téarma.

$$\int x\mathrm{d}x+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

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}.

$$\frac{x^{2}}{2}+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

Athscríobh \sqrt[3]{x} mar x^{\frac{1}{3}}. Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{\frac{1}{3}}\mathrm{d}x le \frac{x^{\frac{4}{3}}}{\frac{4}{3}}. Simpligh.

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

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

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

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

$$\frac{2^{2}}{2}+\frac{3}{4}\times 2^{\frac{4}{3}}-2^{-1}-\left(\frac{1^{2}}{2}+\frac{3}{4}\times 1^{\frac{4}{3}}-1^{-1}\right)$$

Simpligh.

$$\frac{5}{4}+\frac{3\sqrt[3]{2}}{2}$$