a) Ta có: \(\displaystyle x - 1 + \frac{{\ln x}}{x} = x - 1\)\(\displaystyle \Leftrightarrow \frac{{\ln x}}{x} = 0 \Leftrightarrow x = 1\).
Khi đó \(\displaystyle S = \int\limits_1^e {\left| {x - 1 + \frac{{\ln x}}{x} - x + 1} \right|dx} \) \(\displaystyle = \int\limits_1^e {\left| {\frac{{\ln x}}{x}} \right|dx} \) \(\displaystyle = \int\limits_1^e {\frac{{\ln x}}{x}dx} \) \(\displaystyle = \int\limits_1^e {\ln xd\left( {\ln x} \right)} \) \(\displaystyle = \left. {\frac{{{{\ln }^2}x}}{2}} \right|_1^e = \frac{1}{2}\)
b) Ta có: \(\displaystyle {x^3} - {x^2} = \frac{1}{9}\left( {x - 1} \right)\) \(\displaystyle \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - \frac{1}{9}} \right) = 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = \frac{1}{3}\\x = - \frac{1}{3}\end{array} \right.\)
Khi đó:
\(\displaystyle S = \int\limits_{ - \frac{1}{3}}^1 {\left| {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right|dx} \)\(\displaystyle = \int\limits_{ - \frac{1}{3}}^{\frac{1}{3}} {\left| {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right|dx} \) \(\displaystyle + \int\limits_{\frac{1}{3}}^1 {\left| {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right|dx} \)
\(\displaystyle = \left| {\int\limits_{ - \frac{1}{3}}^{\frac{1}{3}} {\left[ {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right]dx} } \right|\) \(\displaystyle + \left| {\int\limits_{\frac{1}{3}}^1 {\left[ {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right]dx} } \right|\)
\(\displaystyle = \left| {\left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^3}}}{3} - \frac{1}{9}.\frac{{{x^2}}}{2} + \frac{1}{9}x} \right)} \right|_{ - \frac{1}{3}}^{\frac{1}{3}}} \right|\) \(\displaystyle + \left| {\left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^3}}}{3} - \frac{1}{9}.\frac{{{x^2}}}{2} + \frac{1}{9}x} \right)} \right|_{\frac{1}{3}}^1} \right|\)
\(\displaystyle = \left| {\frac{7}{{324}} + \frac{1}{{36}}} \right| + \left| { - \frac{1}{{36}} - \frac{7}{{324}}} \right| = \frac{8}{{81}}\)
