Bài tập 3.36 trang 179 SBT Toán 12

a) Ta có:

\(\begin{array}{l}
x + \sin x = x \Leftrightarrow \sin x = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{x = \pi }
\end{array}} \right.
\end{array}\)

Khi đó:

\(\begin{array}{l}
{S_1} = \int \limits_0^\pi  \left| {x + \sin x - x} \right|dx\\
 = \int \limits_0^\pi  \left| {\sin x} \right|dx = \int \limits_0^\pi  \sin xdx\\
 =  - \left. {\cos x} \right|_0^\pi \\
 =  - \cos \pi  + \cos 0 = 1 + 1 = 2
\end{array}\)

\(\begin{array}{l}
{S_2} = \int \limits_\pi ^{2\pi } \left| {x + \sin x - x} \right|dx\\
 = \int \limits_\pi ^{2\pi } \left| {\sin x} \right|dx = \int \limits_\pi ^{2\pi } \left( { - \sin x} \right)dx\\
 = \left. {\cos x} \right|_\pi ^{2\pi }\\
 = \cos 2\pi  - \cos \pi  = 1 + 1 = 2
\end{array}\)

Do đó S₁ = S₂.

b) \({S_1} = \int \limits_0^\pi  \left| {\sin x} \right|dx = \int \limits_0^\pi  \sin xdx\)

\(\begin{array}{l}
 =  - \left. {\cos x} \right|_0^\pi \\
 =  - \cos \pi  + \cos 0 = 1 + 1 = 2
\end{array}\)

\(\begin{array}{l}
{S_2} = \int \limits_0^\pi  \left| {\cos x} \right|dx\\
 = \int \limits_0^{\frac{\pi }{2}} \left| {\cos x} \right|dx + \int \limits_{\frac{\pi }{2}}^\pi  \left| {\cos x} \right|dx\\
 = \int \limits_0^{\frac{\pi }{2}} \cos xdx - \int \limits_{\frac{\pi }{2}}^\pi  \cos xdx\\
 = \left. {\sin x} \right|_0^{\frac{\pi }{2}} - \left. {\sin x} \right|_{\frac{\pi }{2}}^\pi \\
 = \sin \frac{\pi }{2} - \sin 0 - \sin \pi  + \sin \frac{\pi }{2}\\
 = 1 - 0 - 0 + 1 = 2
\end{array}\)

Do đó S₁ = S₂.

c) Ta có: 

\(\begin{array}{l}
\sqrt x  = {x^2} \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x = {x^4}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x({x^3} - 1) = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}\)

Khi đó

\(\begin{array}{l}
{S_1} = \int \limits_0^1 \left| {\sqrt x  - {x^2}} \right|dx\\
 = \left| {\int \limits_0^1 \left( {\sqrt x  - {x^2}} \right)dx} \right|\\
 = \left| {\left. {\left( {\frac{2}{3}{x^{\frac{3}{2}}} - \frac{{{x^3}}}{3}} \right)} \right|_0^1} \right| = \left| {\frac{2}{3} - \frac{1}{3}} \right| = \frac{1}{3}
\end{array}\)

\(\begin{array}{l}
\sqrt {1 - {x^2}}  = 1 - x\\
 \Leftrightarrow \left\{ \begin{array}{l}
1 - x \ge 0\\
1 - {x^2} = {(1 - x)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
1 - {x^2} = 1 - 2x + {x^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
2{x^2} - 2x = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}\)

Khi đó 

\(\begin{array}{l}
{S_2} = \int \limits_0^1 \left| {\sqrt {1 - {x^2}}  - \left( {1 - x} \right)} \right|dx\\
 = \int \limits_0^1 \left| {\sqrt {1 - {x^2}}  - 1 + x} \right|dx\\
 = \left| {\int \limits_0^1 \left( {\sqrt {1 - {x^2}}  - 1 + x} \right)dx} \right|\\
 = \left| {\int \limits_0^1 \sqrt {1 - {x^2}} dx - \int \limits_0^1 dx + \int \limits_0^1 xdx} \right|\\
 = \left| {\int \limits_0^1 \sqrt {1 - {x^2}} dx - 1 + \frac{1}{2}} \right| = \left| {I - \frac{1}{2}} \right|\\

\end{array}\)

Tính \(I = \int \limits_0^1 \sqrt {1 - {x^2}} dx\)

Đặt \(x = \sin t \Rightarrow dx = \cos tdt\)

\(\begin{array}{l}
 \Rightarrow I = \int \limits_0^{\frac{\pi }{2}} \sqrt {1 - {{\sin }^2}t} .\cos tdt\\
 = \int \limits_0^{\frac{\pi }{2}} {\cos ^2}tdt = \frac{1}{2}\int \limits_0^{\frac{\pi }{2}} \left( {1 + \cos 2t} \right)dt\\
 = \frac{1}{2}\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}} = \frac{1}{2}.\frac{\pi }{2} = \frac{\pi }{4}
\end{array}\)

Do đó S₁ ≠ S₂.

-- Mod Toán 12 HỌC247