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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[I\] is the greatest of the definite integrals \[{{I}_{1}}=\int_{0}^{1}{{{e}^{-x}}{{\cos }^{2}}x\,dx},\] \[{{I}_{2}}=\int_{0}^{1}{{{e}^{-{{x}^{2}}}}}{{\cos }^{2}}x\,dx\] \[{{I}_{3}}=\int_{0}^{1}{{{e}^{-{{x}^{2}}}}dx},\] \[{{I}_{4}}=\int_{0}^{1}{{{e}^{-{{x}^{2}}/2}}dx},\] Then

    A) \[I={{I}_{1}}\]                      

    B) \[I={{I}_{2}}\]

    C) \[I={{I}_{3}}\]                      

    D) \[I={{I}_{4}}\]

    Correct Answer: D

    Solution :

    • For\[0<x<1\], we have \[\frac{1}{2}{{x}^{2}}<{{x}^{2}}<x\]                   
    • \[\Rightarrow -{{x}^{2}}>-x,\]so that \[{{e}^{-{{x}^{2}}}}<{{e}^{-x}}\],                   
    • Hence \[\int_{0}^{1}{{{e}^{-{{x}^{2}}}}{{\cos }^{2}}x\,dx}>\int_{0}^{1}{\,\,{{e}^{-x}}{{\cos }^{2}}x\,dx}\].                   
    • Also \[{{\cos }^{2}}x\le 1\]                   
    • Therefore \[\int_{0}^{1}{{{e}^{-{{x}^{2}}}}{{\cos }^{2}}xdx\le \int_{0}^{1}{{{e}^{-{{x}^{2}}}}dx<\int_{0}^{1}{{{e}^{-{{x}^{2}}/2}}dx={{I}_{4}}}}}\]                   
    • Hence \[{{I}_{4}}\]is the greatest integral.


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