Concept:

Trapezoidal rule is given by:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{\;}} \ldots } \right)} \right]\)

\({\rm{Number\;of\;intervals(n)}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{\;}}\)

where b is the upper limit, a is the lower limit, h is the step size.

Calculation:

From question

\(I = \mathop \smallint \nolimits_0^4 2.f\left( x \right).dx\)

Let f(x) = 2.f(x)

∴ \(f\left( x \right) = \frac{{f\left( x \right)}}{2}\)

\(I = \mathop \smallint \nolimits_0^4 \frac{{f\left( x \right)}}{2} \cdot dx = \frac{1}{2}\mathop \smallint \nolimits_0^4 f\left( x \right) \cdot dx\)

From the formula:

\(\mathop \smallint \nolimits_0^4 f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_4}} \right) + 2\left( {{y_1} + {y_2} + {y_3}} \right)} \right]\;\)

\(= \frac{{1}}{2}\left[ {\left( {5 + 7} \right) + 2\left( {2 + 1 + 3} \right)} \right]\)

∴ \(I = \frac{{24}}{2} = 12\)