Correct Answer - Option 4 : P-3; Q-1; R-2
Explanation:
\({\rm{Number\;of\;intervals}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{\;}}\)
where,
b is the upper limit, a is the lower limit, h is the step size
According to the trapezoidal rule
\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{\;}} \ldots } \right)} \right]\)
It fits for 1-degree polynomial.
According to Simpson's 1/3 rule
\({\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{h}{3}\left[ {\left( {{y_o} + {y_n}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots } \right)} \right] + 4\left[ {{y_1} + {y_3} + {y_5} + \ldots } \right]}\)
It fits for 2-degree polynomial.
According to Simpson's 3/8 rule
\(\mathop \smallint \limits_a^b ydx = \frac{{3h}}{8}\left[ {({y_0}+y_n) + 3({y_1} + {y_2}+{y_4} +{y_5}..)+ 2({y_3} + {y_6} + {y_9}..)} \right]\)
It fits for 3-degree polynomial.
