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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    \[\left[ \sum\limits_{n=1}^{10}{\int_{-2n-1}^{2n}{{{\sin }^{27}}x\,dx}} \right]+\left[ \sum\limits_{n=1}^{10}{\int_{2n}^{2n+1}{{{\sin }^{27}}x\,dx}} \right]\] equals [MP PET 2002]

    A) \[{{27}^{2}}\]                       

    B) \[-54\]

    C) 36  

    D) 0

    Correct Answer: D

    Solution :

    • \[\sum\limits_{n=1}^{10}{\int_{-2n-1}^{2n}{{{\sin }^{27}}x\,dx+}}\sum\limits_{n=1}^{10}{\int_{2n}^{2n+1}{{{\sin }^{27}}x\,dx}}\]           
    • In the first summation put \[x=-t\]           
    • \[\sum\limits_{n=1}^{10}{{{\sin }^{27}}(-t)(-dt)+\sum\limits_{n=1}^{10}{\int_{2n}^{2n+1}{{{\sin }^{27}}x\,dx}}}\]                   
    • \[=-\sum\limits_{n=1}^{10}{\int_{2n}^{2n+1}{{{\sin }^{27}}}x\,dx+}\sum\limits_{n=1}^{10}{\int_{2n}^{2n+1}{\,{{\sin }^{27}}}x\,dx=0}\].


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