JCECE Engineering JCECE Engineering Solved Paper-2010

  • question_answer
    With the help of trapezoidal rule for numerical integration and the following table.
    \[x\] \[0\] \[0.25\] \[0.50\] \[0.75\] \[1\]
    \[f(x)\] \[0\] \[0.625\] \[0.2500\] \[0.5625\] \[1\]
    The value of\[\int_{0}^{1}{f(x)}\,\,dx\]is

    A) \[0.35342\]                        

    B) \[0.34375\]

    C) \[0.34457\]                        

    D) \[0.33334\]

    Correct Answer: B

    Solution :

    We know, the trapezoidal rule is                 \[\int_{a}^{b}{f(x)}dx\]                 \[=\frac{h}{2}[({{y}_{0}}+{{y}_{n}})+2({{y}_{1}}+{{y}_{2}}+...+{{y}_{n}})]\] But, we have\[a=0,\,\,b=1,\,\,h=0.25,\,\,n=5\] \[\therefore \]\[\int_{a}^{b}{f(x)}dx\] \[=\frac{0.25}{2}[(0.1+2(0.0625+0.25+0.5625)]\] \[=\frac{0.25}{2}[1+1.75]\] \[=0.34375\]


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