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JEE Main & Advanced Mathematics Definite Integration Question Bank Self Evaluation Test - Application of Integrals

  • question_answer
    If \[y=f(x)\] makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then \[\int\limits_{0}^{2}{xf'(x)dx}\] is

    A) 3/2

    B) 1

    C) 5/4

    D) -3/4

    Correct Answer: D

    Solution :

    [d] We have \[\int\limits_{0}^{2}{f(x)dx=\frac{3}{4}};\] Now, \[\int\limits_{0}^{2}{xf'(x)dx=\,\,x\int\limits_{0}^{2}{f'(x)dx}-\int\limits_{0}^{2}{f(x)dx}}\] \[=[xf(x)]_{0}^{2}-\frac{3}{4}=2f(2)-\frac{3}{4}\] \[=\,\,\,0-\frac{3}{4}(\because f(2)=0)=-\frac{3}{4}.\]


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