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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    The value of the definite integral \[\int_{0}^{1}{\frac{x\,dx}{{{x}^{3}}+16}}\] lies in the interval \[[a,\,\,b].\] The smallest such interval is

    A) \[\left[ 0,\,\,\frac{1}{17} \right]\]                                   

    B) [0, 1]

    C) \[\left[ 0,\,\,\frac{1}{27} \right]\]                                   

    D) None of these

    Correct Answer: A

    Solution :

    • The function \[f(x)=\frac{x}{{{x}^{3}}+16}\]is an increasing function, so Min\[f(x)=f(0)=0\]and Max\[f(x)=f(1)=\frac{1}{17}\]           
    • Therefore by the property \[m(b-a)\le \int_{a}^{b}{f(x)dx\le M(b-a)}\]           
    • (where m and M are the smallest and greatest values of function)           
    • Þ \[0\le \int_{0}^{1}{\frac{x}{{{x}^{3}}+16}dx\le \frac{1}{17}}\].


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