JEE Main & Advanced Mathematics Definite Integration Question Bank Properties of Definite Integration

  • question_answer
    If \[f(x)\] is a continuous periodic function with period \[T,\] then the integral \[I=\int_{a}^{a+T}{f(x)\,dx}\] is

    A)                 Equal to \[2a\]  

    B)                 Equal to \[3a\]

    C)                 Independent of \[a\]    

    D)                 None of these

    Correct Answer: C

    Solution :

               Consider the function \[g(a)=\int_{a}^{a+T}{f(x)dx}\]                    \[=\int_{a}^{0}{f(x)dx+\int_{0}^{T}{f(x)dx+\int_{T}^{a+T}{\,\,f(x)dx}}}\]            Putting \[x-T=y\] in last integral, we get \[\int_{T}^{a+T}{f(x)dx=\int_{0}^{a}{f(y+T)dy=\int_{0}^{a}{f(y)dy}}}\]            Þ \[g(a)=\int_{a}^{0}{f(x)dx+\int_{0}^{1}{f(x)dx+\int_{0}^{a}{f(x)dx}}}\]\[=\int_{0}^{T}{f(x)dx}\]                                 Hence g is independent of a.


You need to login to perform this action.
You will be redirected in 3 sec spinner