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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    Let \[\int_{0}^{1}{f(x)\,dx=1,}\] \[\int_{0}^{1}{x\,f(x)\,dx=a}\] and \[\int_{0}^{1}{{{x}^{2}}f(x)\,dx={{a}^{2}},}\] then the value of \[\int_{0}^{1}{{{(x-a)}^{2}}f(x)\,dx=}\] [IIT 1990]

    A) 0    

    B) \[{{a}^{2}}\]

    C) \[{{a}^{2}}-1\]                     

    D) \[{{a}^{2}}-2a+2\]

    Correct Answer: A

    Solution :

    • \[\int_{0}^{1}{{{(x-a)}^{2}}f(x)dx=\int_{0}^{1}{{{x}^{2}}f(x)dx+{{a}^{2}}\int_{0}^{1}{f(x)dx}}}-\int_{0}^{1}{2axf(x)dx}\]                   
    • =\[{{a}^{2}}+{{a}^{2}}-2a\times a=0\].


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