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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 value of the integral \[\sum\limits_{k=1}^{n}{\int_{0}^{1}{f(k-1+x)\,dx}}\] is

    A)                 \[\int_{0}^{1}{f(x)\,dx}\]             

    B)                 \[\int_{0}^{2}{f(x)\,dx}\]

    C)                 \[\int_{0}^{n}{f(x)\,dx}\]            

    D)                 \[n\int_{0}^{1}{f(x)\,dx}\]

    Correct Answer: C

    Solution :

                       Let \[I=\int_{0}^{1}{f(k-1+x)dx}\]                    Þ \[I=\int_{k-1}^{k}{f(t)\,dt,}\]where \[t=k-1+x\]Þ \[I=\int_{k-1}^{k}{f(x)dx}\]                \[\therefore \,\,\,\sum\limits_{k=1}^{n}{\int_{k-1}^{k}{f(x)dx=\int_{0}^{1}{f(x)dx+\int_{1}^{2}{f(x)dx+.....+\int_{n-1}^{n}{f(x)dx}}}}}\]                                                            \[=\int_{0}^{n}{f(x)\,dx}\].


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