Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    The approximate value of\[\int_{1}^{5}{{{x}^{2}}dx}\]using trapezoidal rule with n = 4 is

    A)  41                                         

    B)  41.5

    C)  41.75   

    D)         42

    Correct Answer: D

    Solution :

    \[\int_{1}^{5}{{{x}^{2}}dx}\] \[=\frac{4}{4}\left[ \frac{1}{2}f(1)+f(2)=f(3)+f(4)+\frac{1}{2}f(5) \right]\] \[=\left[ \frac{1}{2}+4+9+16+\frac{25}{2} \right]=42\][using\[f(x)={{x}^{2}}\]]


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