∫ (from 1 to 4) of (x^2+1/x^2) wrt x eching

$$\int _ { 1 } ^ { 4 } ( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } ) d x$$

Baholash

$$\frac{87}{4}=21,75$$

Yechim qadamlari

Avval noaniq integralni baholang.

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

Summani muddatma-muddat integratsiya qiling.

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

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{2}\mathrm{d}x integralni \frac{x^{3}}{3} bilan almashtiring.

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

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int \frac{1}{x^{2}}\mathrm{d}x integralni -\frac{1}{x} bilan almashtiring.

$$\frac{x^{3}}{3}-\frac{1}{x}$$

Xos integral bu integral hisoblashning yuqori chegarasida hisoblangan ifodaning boshlangʻich holatidan chiqarib tashlagan holda integral hisoblashning quyi chegarasida hisoblangan ifodaning boshlangʻich holatidir.

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

Qisqartirish.

$$\frac{87}{4}$$