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Câu hỏi: Nhờ ý nghĩa hình học của tích phân, hãy tìm khẳng định sai trong các khẳng định sau:
A. \(\displaystyle \int\limits_0^1 {\ln \left( {1 + x} \right)dx} > \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} \)
B. \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}xdx} < \int\limits_0^{\frac{\pi }{4}} {\sin 2xdx} \)
C. \(\displaystyle \int\limits_0^1 {{e^{ - x}}dx} > \int\limits_0^1 {{{\left( {\frac{{1 - x}}{{1 + x}}} \right)}^2}dx} \)
D. \(\displaystyle \int\limits_0^1 {{e^{ - {x^2}}}dx} > \int\limits_0^1 {{e^{ - {x^3}}}dx} \)
A. \(\displaystyle \int\limits_0^1 {\ln \left( {1 + x} \right)dx} > \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} \)
B. \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}xdx} < \int\limits_0^{\frac{\pi }{4}} {\sin 2xdx} \)
C. \(\displaystyle \int\limits_0^1 {{e^{ - x}}dx} > \int\limits_0^1 {{{\left( {\frac{{1 - x}}{{1 + x}}} \right)}^2}dx} \)
D. \(\displaystyle \int\limits_0^1 {{e^{ - {x^2}}}dx} > \int\limits_0^1 {{e^{ - {x^3}}}dx} \)
Phương pháp giải
Sử dụng ý nghĩa hình học của tích phân: Nếu \(\displaystyle f\left( x \right) \ge 0,\forall x \in \left[ {a; b} \right]\) thì \(\displaystyle S = \int\limits_a^b {f\left( x \right)dx} > 0\).
Lời giải chi tiết
:
Xét \(\displaystyle I = \int\limits_0^1 {\ln \left( {1 + x} \right)dx} - \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} \) \(\displaystyle = \int\limits_0^1 {\left( {\ln \left( {1 + x} \right) - \frac{{x - 1}}{{e - 1}}} \right)dx} \)
Dễ thấy trong \(\displaystyle \left[ {0; 1} \right]\) thì:
\(\displaystyle \ln \left( {x + 1} \right) \ge 0 \ge \frac{{x - 1}}{{e - 1}}\)\(\displaystyle \Rightarrow \ln \left( {x + 1} \right) - \frac{{x - 1}}{{e - 1}} \ge 0\)\(\displaystyle \Rightarrow \int\limits_0^1 {\left( {\ln \left( {1 + x} \right) - \frac{{x - 1}}{{e - 1}}} \right)dx} > 0\)
\(\displaystyle \Rightarrow \int\limits_0^1 {\ln \left( {1 + x} \right)dx} - \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} > 0\) \(\displaystyle \Leftrightarrow \int\limits_0^1 {\ln \left( {1 + x} \right)dx} > \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} \) hay A đúng.
: Xét \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}xdx} - \int\limits_0^{\frac{\pi }{4}} {\sin 2xdx} \)\(\displaystyle = \int\limits_0^{\frac{\pi }{4}} {\left( {{{\sin }^2}x - \sin 2x} \right)dx} \) \(\displaystyle = \int\limits_0^{\frac{\pi }{4}} {\sin x\left( {\sin x - 2\cos x} \right)dx} \)
Trong đoạn \(\displaystyle \left[ {0;\frac{\pi }{4}} \right]\) thì \(\displaystyle 0 \le \sin x \le \frac{{\sqrt 2 }}{2} \le \cos x \le 1\) \(\displaystyle \Rightarrow \sin x - 2\cos x < 0\)
\(\displaystyle \Rightarrow \sin x\left( {\sin x - 2\cos x} \right) \le 0\) \(\displaystyle \Rightarrow \int\limits_0^{\frac{\pi }{4}} {\sin x\left( {\sin x - 2\cos x} \right)dx} < 0\)\(\displaystyle \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}xdx} < \int\limits_0^{\frac{\pi }{4}} {\sin 2xdx} \) hay B đúng.
: Xét \(\displaystyle \int\limits_0^1 {{e^{ - {x^2}}}dx} - \int\limits_0^1 {{e^{ - {x^3}}}dx} \)\(\displaystyle = \int\limits_0^1 {\left( {{e^{ - {x^2}}} - {e^{ - {x^3}}}} \right)dx} \)
Trong đoạn \(\displaystyle \left[ {0; 1} \right]\) thì \(\displaystyle {x^2} \ge {x^3} \Rightarrow - {x^2} \le - {x^3}\) \(\displaystyle \Rightarrow {e^{ - {x^2}}} \le {e^{ - {x^3}}} \Rightarrow {e^{ - {x^2}}} - {e^{ - {x^3}}} \le 0\)
\(\displaystyle \Rightarrow \int\limits_0^1 {\left( {{e^{ - {x^2}}} - {e^{ - {x^3}}}} \right)dx} < 0\)\(\displaystyle \Leftrightarrow \int\limits_0^1 {{e^{ - {x^2}}}dx} < \int\limits_0^1 {{e^{ - {x^3}}}dx} \) hay D sai.
Sử dụng ý nghĩa hình học của tích phân: Nếu \(\displaystyle f\left( x \right) \ge 0,\forall x \in \left[ {a; b} \right]\) thì \(\displaystyle S = \int\limits_a^b {f\left( x \right)dx} > 0\).
Lời giải chi tiết
:
Xét \(\displaystyle I = \int\limits_0^1 {\ln \left( {1 + x} \right)dx} - \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} \) \(\displaystyle = \int\limits_0^1 {\left( {\ln \left( {1 + x} \right) - \frac{{x - 1}}{{e - 1}}} \right)dx} \)
Dễ thấy trong \(\displaystyle \left[ {0; 1} \right]\) thì:
\(\displaystyle \ln \left( {x + 1} \right) \ge 0 \ge \frac{{x - 1}}{{e - 1}}\)\(\displaystyle \Rightarrow \ln \left( {x + 1} \right) - \frac{{x - 1}}{{e - 1}} \ge 0\)\(\displaystyle \Rightarrow \int\limits_0^1 {\left( {\ln \left( {1 + x} \right) - \frac{{x - 1}}{{e - 1}}} \right)dx} > 0\)
\(\displaystyle \Rightarrow \int\limits_0^1 {\ln \left( {1 + x} \right)dx} - \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} > 0\) \(\displaystyle \Leftrightarrow \int\limits_0^1 {\ln \left( {1 + x} \right)dx} > \int\limits_0^1 {\frac{{x - 1}}{{e - 1}}dx} \) hay A đúng.
: Xét \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}xdx} - \int\limits_0^{\frac{\pi }{4}} {\sin 2xdx} \)\(\displaystyle = \int\limits_0^{\frac{\pi }{4}} {\left( {{{\sin }^2}x - \sin 2x} \right)dx} \) \(\displaystyle = \int\limits_0^{\frac{\pi }{4}} {\sin x\left( {\sin x - 2\cos x} \right)dx} \)
Trong đoạn \(\displaystyle \left[ {0;\frac{\pi }{4}} \right]\) thì \(\displaystyle 0 \le \sin x \le \frac{{\sqrt 2 }}{2} \le \cos x \le 1\) \(\displaystyle \Rightarrow \sin x - 2\cos x < 0\)
\(\displaystyle \Rightarrow \sin x\left( {\sin x - 2\cos x} \right) \le 0\) \(\displaystyle \Rightarrow \int\limits_0^{\frac{\pi }{4}} {\sin x\left( {\sin x - 2\cos x} \right)dx} < 0\)\(\displaystyle \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}xdx} < \int\limits_0^{\frac{\pi }{4}} {\sin 2xdx} \) hay B đúng.
: Xét \(\displaystyle \int\limits_0^1 {{e^{ - {x^2}}}dx} - \int\limits_0^1 {{e^{ - {x^3}}}dx} \)\(\displaystyle = \int\limits_0^1 {\left( {{e^{ - {x^2}}} - {e^{ - {x^3}}}} \right)dx} \)
Trong đoạn \(\displaystyle \left[ {0; 1} \right]\) thì \(\displaystyle {x^2} \ge {x^3} \Rightarrow - {x^2} \le - {x^3}\) \(\displaystyle \Rightarrow {e^{ - {x^2}}} \le {e^{ - {x^3}}} \Rightarrow {e^{ - {x^2}}} - {e^{ - {x^3}}} \le 0\)
\(\displaystyle \Rightarrow \int\limits_0^1 {\left( {{e^{ - {x^2}}} - {e^{ - {x^3}}}} \right)dx} < 0\)\(\displaystyle \Leftrightarrow \int\limits_0^1 {{e^{ - {x^2}}}dx} < \int\limits_0^1 {{e^{ - {x^3}}}dx} \) hay D sai.
Đáp án A.