a) Ta có: \(\displaystyle x + \sin x = x \Leftrightarrow \sin x = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = \pi \end{array} \right.\)
Khi đó \(\displaystyle {S_1} = \int\limits_0^\pi {\left| {x + \sin x - x} \right|dx} \) \(\displaystyle = \int\limits_0^\pi {\left| {\sin x} \right|dx} \) \(\displaystyle = \int\limits_0^\pi {\sin xdx} = - \left. {\cos x} \right|_0^\pi \) \(\displaystyle = - \cos \pi + \cos 0 = 1 + 1 = 2\)
\(\displaystyle {S_2} = \int\limits_\pi ^{2\pi } {\left| {x + \sin x - x} \right|dx} \) \(\displaystyle = \int\limits_\pi ^{2\pi } {\left| {\sin x} \right|dx} \) \(\displaystyle = \int\limits_\pi ^{2\pi } {\left( { - \sin x} \right)dx} = \left. {\cos x} \right|_\pi ^{2\pi }\) \(\displaystyle = \cos 2\pi - \cos \pi = 1 + 1 = 2\)
Do đó \(\displaystyle {S_1} = {S_2}\).
b) \(\displaystyle {S_1} = \int\limits_0^\pi {\left| {\sin x} \right|dx} = \int\limits_0^\pi {\sin xdx} \) \(\displaystyle = - \left. {\cos x} \right|_0^\pi \)\(\displaystyle = - \cos \pi + \cos 0 = 1 + 1 = 2\)
\(\displaystyle {S_2} = \int\limits_0^\pi {\left| {\cos x} \right|dx} \) \(\displaystyle = \int\limits_0^{\frac{\pi }{2}} {\left| {\cos x} \right|dx} + \int\limits_{\frac{\pi }{2}}^\pi {\left| {\cos x} \right|dx} \) \(\displaystyle = \int\limits_0^{\frac{\pi }{2}} {\cos xdx} - \int\limits_{\frac{\pi }{2}}^\pi {\cos xdx} \) \(\displaystyle = \left. {\sin x} \right|_0^{\frac{\pi }{2}} - \left. {\sin x} \right|_{\frac{\pi }{2}}^\pi \)
\(\displaystyle = \sin \frac{\pi }{2} - \sin 0 - \sin \pi + \sin \frac{\pi }{2}\) \(\displaystyle = 1 - 0 - 0 + 1 = 2\)
Do đó \(\displaystyle {S_1} = {S_2}\).
c) Ta có: \(\displaystyle \sqrt x = {x^2} \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x = {x^4}\end{array} \right.\) \(\displaystyle \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x\left( {{x^3} - 1} \right) = 0\end{array} \right.\) \(\displaystyle \left[ \begin{array}{l}x = 0\\x = 1\end{array} \right.\)
Khi đó \(\displaystyle {S_1} = \int\limits_0^1 {\left| {\sqrt x - {x^2}} \right|dx} \) \(\displaystyle = \left| {\int\limits_0^1 {\left( {\sqrt x - {x^2}} \right)dx} } \right|\) \(\displaystyle = \left| {\left. {\left( {\frac{2}{3}{x^{\frac{3}{2}}} - \frac{{{x^3}}}{3}} \right)} \right|_0^1} \right| = \left| {\frac{2}{3} - \frac{1}{3}} \right| = \frac{1}{3}\)
\(\displaystyle \sqrt {1 - {x^2}} = 1 - x\) \(\displaystyle \Leftrightarrow \left\{ \begin{array}{l}1 - x \ge 0\\1 - {x^2} = {\left( {1 - x} \right)^2}\end{array} \right.\) \(\displaystyle \Leftrightarrow \left\{ \begin{array}{l}x \le 1\\1 - {x^2} = 1 - 2x + {x^2}\end{array} \right.\)
\(\displaystyle \Leftrightarrow \left\{ \begin{array}{l}x \le 1\\2{x^2} - 2x = 0\end{array} \right.\) \(\displaystyle \Leftrightarrow \left\{ \begin{array}{l}x \le 1\\\left[ \begin{array}{l}x = 0\\x = 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 1\end{array} \right.\)
Khi đó \(\displaystyle {S_2} = \int\limits_0^1 {\left| {\sqrt {1 - {x^2}} - \left( {1 - x} \right)} \right|dx} \) \(\displaystyle = \int\limits_0^1 {\left| {\sqrt {1 - {x^2}} - 1 + x} \right|dx} \) \(\displaystyle = \left| {\int\limits_0^1 {\left( {\sqrt {1 - {x^2}} - 1 + x} \right)dx} } \right|\)
\(\displaystyle = \left| {\int\limits_0^1 {\sqrt {1 - {x^2}} dx} - \int\limits_0^1 {dx} + \int\limits_0^1 {xdx} } \right|\) \(\displaystyle = \left| {\int\limits_0^1 {\sqrt {1 - {x^2}} dx} - 1 + \frac{1}{2}} \right|\) \(\displaystyle = \left| {I - \frac{1}{2}} \right|\)
Tính \(\displaystyle I = \int\limits_0^1 {\sqrt {1 - {x^2}} dx} \).
Đặt \(\displaystyle x = \sin t \Rightarrow dx = \cos tdt\) \(\displaystyle \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\sqrt {1 - {{\sin }^2}t} .\cos tdt} \) \(\displaystyle = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} \)
\(\displaystyle = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2t} \right)dt} \) \(\displaystyle = \frac{1}{2}\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}}\) \(\displaystyle = \frac{1}{2}.\frac{\pi }{2} = \frac{\pi }{4}\)
Do đó \(\displaystyle {S_1} \ne {S_2}\).
