Bài tập 3.46 trang 181 SBT Toán 12

a) Ta có: 

\(\begin{array}{l}
x - 1 + \frac{{\ln x}}{x} = x - 1\\
 \Leftrightarrow \frac{{\ln x}}{x} = 0 \Leftrightarrow x = 1
\end{array}\)

Khi đó${\displaystyle \frac{\mathrm{}}{}}$

\(\begin{array}{l}
S = \int \limits_1^e \left| {x - 1 + \frac{{\ln x}}{x} - x + 1} \right|dx\\
 = \int \limits_1^e \left| {\frac{{\ln x}}{x}} \right|dx = \int \limits_1^e \frac{{\ln x}}{x}dx\\
 = \int \limits_1^e \ln xd\left( {\ln x} \right) = \left. {\frac{{{{\ln }^2}x}}{2}} \right|_1^e = \frac{1}{2}
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
{x^3} - {x^2} = \frac{1}{9}\left( {x - 1} \right)\\
 \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - \frac{1}{9}} \right) = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 1}\\
{x = \frac{1}{3}}\\
{x =  - \frac{1}{3}}
\end{array}} \right.
\end{array}\)

Khi đó:

\(\begin{array}{l}
S = \int \limits_{ - \frac{1}{3}}^1 \left| {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right|dx\\
 = \int \limits_{ - \frac{1}{3}}^{\frac{1}{3}} \left| {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right|dx\\
 + \int \limits_{\frac{1}{3}}^1 \left| {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right|dx\\
 = \left| {\int \limits_{ - \frac{1}{3}}^{\frac{1}{3}} \left[ {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right]dx} \right|\\
 + \left| {\int \limits_{\frac{1}{3}}^1 \left[ {{x^3} - {x^2} - \frac{1}{9}\left( {x - 1} \right)} \right]dx} \right|\\
 = \left| {\left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^3}}}{3} - \frac{1}{9}.\frac{{{x^2}}}{2} + \frac{1}{9}x} \right)} \right|_{\frac{{ - 1}}{3}}^{\frac{1}{3}}} \right|\\
 + \left| {\left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^3}}}{3} - \frac{1}{9}.\frac{{{x^2}}}{2} + \frac{1}{9}x} \right)} \right|_{\frac{1}{3}}^1} \right|\\
 = \left| {\frac{7}{{324}} + \frac{1}{{36}}} \right| + \left| { - \frac{1}{{36}} - \frac{7}{{324}}} \right| = \frac{8}{{81}}
\end{array}\)

-- Mod Toán 12 HỌC247