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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[{{I}_{1}}=\int_{0}^{1}{{{2}^{{{x}^{2}}}}dx,\ }{{I}_{2}}=\int_{0}^{1}{{{2}^{{{x}^{3}}}}dx},\ {{I}_{3}}=\int_{1}^{2}{{{2}^{{{x}^{2}}}}dx}\],\[{{I}_{4}}=\int_{1}^{2}{{{2}^{{{x}^{3}}}}dx}\],  then [AIEEE 2005]

    A) \[{{I}_{3}}={{I}_{4}}\]       

    B) \[{{I}_{3}}>{{I}_{4}}\]

    C) \[{{I}_{2}}>{{I}_{1}}\]       

    D) \[{{I}_{1}}>{{I}_{2}}\]

    Correct Answer: D

    Solution :

    • For \[0<x<1\], we have \[{{x}^{2}}>{{x}^{3}}\]  and for \[1<x<2\], we have \[{{x}^{3}}>{{x}^{2}}\]                   
    • \[\therefore \] \[{{2}^{{{x}^{2}}}}>{{2}^{{{x}^{3}}}}\] for \[0<x<1\] and \[{{2}^{{{x}^{2}}}}<{{2}^{{{x}^{3}}}}\] for \[1<x<2\]                   
    • \[\therefore \]  \[\int_{0}^{1}{{{2}^{{{x}^{2}}}}dx>}\int_{0}^{1}{{{2}^{{{x}^{3}}}}dx}\] and \[\int_{1}^{2}{{{2}^{{{x}^{2}}}}dx<\int_{1}^{2}{{{2}^{{{x}^{3}}}}dx}}\]     \[\therefore \]  \[{{I}_{1}}>{{I}_{2}}\] and \[{{I}_{3}}<{{I}_{4}}\].


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