Tính diện tích hình phẳng giới hạn bởi các đường \(y = {x^2} - x,\,\,x - y + 3 = 0\)

Ta có: \(\mathop {}\limits_{x \to + \infty } \ln 2x = + \infty \Rightarrow x + \ln 2x \sim x\)

Khi đó: \(\int\limits_2^{ + \infty } {\frac{{dx}}{{\sqrt {x + \ln 2x} }}} \sim \int\limits_2^{ + \infty } {\frac{{dx}}{{\sqrt x }}} \)

Mà \(\int\limits_2^{ + \infty } {\frac{{dx}}{{\sqrt x }}} = \mathop {\lim }\limits_{b \to + \infty } \int\limits_2^b {\frac{{dx}}{{\sqrt x }}} = \mathop {\lim }\limits_{b \to + \infty } \left. {2\sqrt x } \right|_2^b = \mathop {\lim }\limits_{b \to + \infty } \left( {2\sqrt b - 2\sqrt 2 } \right) = + \infty \Rightarrow \int\limits_2^{ + \infty } {\frac{{dx}}{{\sqrt x }}} \) phân kỳ.

Vậy \(\int\limits_2^{ + \infty } {\frac{{dx}}{{\sqrt {x + \ln 2x} }}}\) phân kỳ.

Ta có:
\(\int\limits_2^4 {\frac{{dx}}{{\sqrt {x - 2} }}} = \mathop {lim}\limits_{\varepsilon  \to {0^ + }} \int\limits_{2 + \varepsilon }^4 {\frac{{dx}}{{\sqrt {x - 2} }}} = \mathop {lim}\limits_{\varepsilon  \to {0^ + }} \left( {2\sqrt {x - 2} } \right)\left| {_{2 + \varepsilon }^4} \right. = \mathop {lim}\limits_{\varepsilon  \to {0^ + }} \left( {2\sqrt 2  - 2\sqrt {2 + \varepsilon  - 2} } \right) = 2\sqrt 2 } \)