Задача 57862 Помогите...

600eece9a8b7045a0a5301ec

11.02.2021 23:46:35

Помогите

5f3ea7e3faf909182968ddd9

12.02.2021 00:12:59

★

[m] ∫ ^{2,5}_{2}\frac{dx}{(2x-4)^{\frac{1}{2}}}= ∫ ^{2,5}_{2}\frac{dx}{\sqrt{2(x-2)}}=\frac{1}{\sqrt{2}}∫ ^{2,5}_{2}\frac{d(x-2)}{\sqrt{x-2}}

=\frac{1}{\sqrt{2}}\cdot 2\sqrt{x-2}|^{2,5}_{2}=\sqrt{2}\cdot (\sqrt{2,5-2}-\sqrt{2-2})=\sqrt{2}\cdot \sqrt{0,5}=1[/m]