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JEE Main & Advanced Mathematics Definite Integration Question Bank Summation of series by Definite Integration, Gamma function, Leibnitz's rule

  • question_answer
    The least value of the function \[F(x)=\] \[\int_{5\pi /4}^{x}{(3\sin u+4\cos u)\,du}\] on the interval \[\left[ \frac{5\pi }{4},\,\,\frac{4\pi }{3} \right]\] is

    A)                 \[\sqrt{3}+\frac{3}{2}\]

    B)                 \[-2\sqrt{3}+\frac{3}{2}+\frac{1}{\sqrt{2}}\]

    C)                 \[\frac{3}{2}+\frac{1}{\sqrt{2}}\]              

    D)                 None of these

    Correct Answer: B

    Solution :

                       We have \[F'(x)=3\sin x+4\cos x\]                    Since in \[\left[ \frac{5\pi }{4},\frac{4\pi }{3} \right],F'(x)<0,\]so assume the least value at the point \[x=\frac{4\pi }{3}.\]                    Thus, \[f\,\,\left( \frac{4\pi }{3} \right)=\int_{5\pi /4}^{4\pi /3}{\,\,(3\sin u+4\cos u)du}\]                                                           \[=\frac{3}{2}-2\sqrt{3}+\frac{1}{\sqrt{2}}\].


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